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#### Page No 382:

#### Question 1:

How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?

#### Answer:

Balmer series contains wavelengths ranging from 364 nm (for n_{2} = 3) to 655 nm (n_{2}_{}= $\infty $).

So, the given range of wavelength (380−780 nm) lies in the Balmer series.

The wavelength in the Balmer series can be found by

$\frac{1}{\lambda}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{n}^{2}}\right)$

Here, *R* = Rydberg's constant = 1.097×10^{7} m^{$-$1}

The wavelength for the transition from *n* = 3 to *n* = 2 is given by

$\frac{1}{{\lambda}_{1}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{1}=656.3\mathrm{nm}$

The wavelength for the transition from *n* = 4 to *n* = 2 is given by

$\frac{1}{{\lambda}_{2}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{2}=486.1\mathrm{nm}$

The wavelength for the transition from *n* = 5 to *n* = 2 is given by

$\frac{1}{{\lambda}_{3}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{5}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{3}=434.0\mathrm{nm}$

The wavelength for the transition from *n* = 6 to *n* = 2 is given by

$\frac{1}{{\lambda}_{4}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{6}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{4}=410.2\mathrm{nm}$

The wavelength for the transition from *n* = 7 to *n* = 2 is given by

$\frac{1}{{\lambda}_{5}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{7}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{5}=397.0\mathrm{nm}$

Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.

Lyman series contains wavelengths ranging from 91 nm (for n_{2} = 2) to 121 nm (n_{2 }=$\infty $).

So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series.

The wavelength in the Lyman series can be found by

$\frac{1}{\lambda}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)$

The wavelength for the transition from *n* = 2 to *n* = 1 is given by

$\frac{1}{{\lambda}_{1}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{1}=122\mathrm{nm}$

The wavelength for the transition from *n* = 3 to *n* = 1 is given by

$\frac{1}{{\lambda}_{2}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{2}=103\mathrm{nm}$

The wavelength for the transition from *n* = 4 to *n* = 1 is given by

$\frac{1}{{\lambda}_{3}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{4}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{3}=97.3\mathrm{nm}$

The wavelength for the transition from *n* = 5 to *n* = 1 is given by

$\frac{1}{{\lambda}_{4}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{5}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{4}=95.0\mathrm{nm}$

The wavelength for the transition from *n* = 6 to *n* = 1 is given by

$\frac{1}{{\lambda}_{5}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{6}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda}_{5}=93.8\mathrm{nm}$

So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.

#### Page No 382:

#### Question 2:

The first excited energy of a He^{+} ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?

#### Answer:

The energy of hydrogen ion is given by

${E}_{\mathrm{n}}=-\frac{(13.6\mathrm{eV}){Z}^{2}}{{n}^{2}}$

For the first excited state (*n* = 2), the energy of He^{+} ion (with *Z =* 2) will be $-$13.6 eV. This is same as the ground state energy of a hydrogen atom.

Similarly, for all the hydrogen like ions, the energy of the (n $-$ 1)th excited state will be same as the ground state energy of a hydrogen atom if *Z = n.*

#### Page No 382:

#### Question 3:

Which wavelengths will be emitted by a sample of atomic hydrogen gas (in ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?

#### Answer:

As the electron collides, it transfers all its energy to the hydrogen atom.

The excitation energy to raise the electron from the ground state to the nth state is given by

$E=(13.6\mathrm{eV})\times \left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)$

Substituting *n* = 2, we get*E* = 10.2 eV

Substituting *n* = 3, we get*E*' = 12.08 eV

Thus, the atom will be raised to the second excited energy level.

So, when it comes to the ground state, there is transitions from *n* = 3 to *n* = 1.

Therefore, the wavelengths emitted will lie in the Lyman series (infrared region).

#### Page No 382:

#### Question 4:

When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in Lyman series only. Explain.

#### Answer:

White radiations are x-rays that have energy ranging between 5-10 eV.

When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed only in the Lyman series.

At room temperature, almost all the atoms are in ground state.

The minimum energy required for absorption is 10.2 eV (for a transition from n = 1 to n = 2).

The white radiation has photon radiations that have an energy of around 10.2 eV.

So, they are just sufficient to transmit an electron from n = 1 to n = 2 level.

Hence, the absorption lines are observed only in the Lyman series.

#### Page No 383:

#### Question 6:

In which of the following systems will the wavelength corresponding to *n* = 2 to *n* = 1 be minimum?

(a) Hydrogen atom

(b) Deuterium atom

(c) Singly ionized helium

(d) Doubly ionized lithium

#### Answer:

(d) Doubly ionized lithium

The wavelength corresponding the transition from* **n*_{2} to *n*_{1} is given by

$\frac{1}{\lambda}=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

Here,*R* = Rydberg constant*Z* = Atomic number of the ion

From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.

Therefore, the wavelength corresponding to *n* = 2 to *n* = 1 will be minimum in doubly ionized lithium ion because for lithium, *Z* = 3.

#### Page No 383:

#### Question 5:

Balmer series was observed and analysed before the other series. Can you suggest a reason for such an order?

#### Answer:

The Balmer series lies in the visible range. Therefore, it was observed and analysed before the other series. The wavelength range of Balmer series is from 364 nm (for n_{2} =$\infty $) to 655 nm (for n_{2 }=3).

#### Page No 383:

#### Question 6:

What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write E_{n} = E_{1}/*n*^{2}, or *r _{n}* =

*a*

_{0}

*n*

^{2}?

#### Answer:

Energy of n^{th} state of hydrogen is given by

${E}_{\mathrm{n}}=\frac{-13.6}{{n}^{2}}\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

Energy of first excited state (n = 2) of hydrogen, *E*_{1} = $\frac{-13.6}{4}\mathrm{eV}$ = -3.4 eV

This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton so here our refrence point energy is 10 eV. Earlier The energy of first excited state was -3.4 eV when the refrence point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,*E _{1}*

^{,}= -3.4 eV-10 eV = -13.4 eV

We still write E

_{n}= E

_{1}/

*n*

^{2}, or

*r*=

_{n}*a*

_{0}

*n*

^{2}because these formulas are independent of the refrence point enegy.

#### Page No 383:

#### Question 7:

The difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series. Explain.

#### Answer:

The 'series limit' refers to the 'shortest wavelength' (corresponding to the maximum photon energy).

The frequency of the radiation emitted for transition from *n*_{1} to *n*_{2}_{ }is given by

$f=k\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

Here, *k* is a constant.

For the series limit of Lyman series,*n*_{1} = 1*n*_{2} = $\infty $

Frequency, ${f}_{1}=k\left(\frac{1}{{1}^{2}}-\frac{1}{\infty}\right)=k$

For the first line of Lyman series,*n*_{1} = 1*n*_{2} = 2

Frequency, ${f}_{2}=k\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)=\frac{3k}{4}$

For series limit of Balmer series,*n*_{1} = 2*n*_{2} = $\infty $

${f}_{1}=k\left(\frac{1}{{2}^{2}}-\frac{1}{\infty}\right)=\frac{k}{4}$*f*_{1}_{}$-$*f*_{3}_{}= *f*_{2}_{}

Thus, the difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.

#### Page No 383:

#### Question 8:

The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?

#### Answer:

The electron volt is the amount of energy given to an electron in order to move it through the electric potential difference of one volt.

1 eV = 1.6 × 10^{–19} J

The numerical value of ionisation energy in eV is equal to the ionisation potential in volts. The equality does not hold if these quantities are measured in some other units.

#### Page No 383:

#### Question 9:

We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?

#### Answer:

When a photon of energy *(**E*_{2} – *E*_{1} = *hυ)* is incident on an atom in the ground state, the atom in the ground state *E*_{1} may absorb the photon and jump to a higher energy state (*E*_{2}). This process is called stimulated absorption or induced absorption.

Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon.

We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.

#### Page No 383:

#### Question 10:

An atom is in its excited state. Does the probability of its coming to ground state depend on whether the radiation is already present or not? If yes, does it also depend on the wavelength of the radiation present?

#### Answer:

When an atom transits from an excited state to ground state in the presence of an external radiation then it is called as stimulated transition.

When an atom transits from an excited state to ground state on its own then it is called as spontaneous transition.

Ratio of the coefficient for stimulated transition to spontaneous transition is given by

$R=\frac{A}{B\rho \left(\nu \right)}={e}^{\frac{h\nu}{kT}}-1\phantom{\rule{0ex}{0ex}}$

For microwave region

$\nu ={10}^{10}\left(\mathrm{say}\right)\phantom{\rule{0ex}{0ex}}\mathrm{R}={\mathrm{e}}^{\frac{6.6\times {10}^{-34}\times {10}^{10}}{1.38\times {10}^{-23}\times 300}}-1\phantom{\rule{0ex}{0ex}}={\mathrm{e}}^{0.0016}-1\phantom{\rule{0ex}{0ex}}=0.0016$

This implies that stimulated transition dominate in this region.

For visible region

$\nu ={10}^{15}\phantom{\rule{0ex}{0ex}}\Rightarrow R={e}^{160}-1\phantom{\rule{0ex}{0ex}}\Rightarrow R1$

So here spontaneous transition dominate.

#### Page No 383:

#### Question 1:

The minimum orbital angular momentum of the electron in a hydrogen atom is

(a) *h*

(b) *h*/2

(c) *h*/2π

(d) *h*/λ

#### Answer:

(c) *h*/2π

According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of *h*/2π.

∴ ${L}_{\mathrm{n}}=\frac{nh}{2\pi}$

Here,*n* = Principal quantum number

The minimum value of n is 1.

Thus, the minimum value of the orbital angular momentum of the electron in a hydrogen atom is given by

$L=\frac{h}{2\pi}$

#### Page No 383:

#### Question 2:

Three photons coming from excited atomic-hydrogen sample are picked up. Their energies are 12.1 eV, 10.2 eV and 1.9 eV. These photons must come from

(a) a single atom

(b) two atoms

(c) three atoms

(d) either two atoms or three atoms

#### Answer:

(d) either two atoms or three atoms

The energies of the photons emitted can be expressed as follows:

$13.6\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\mathrm{eV}=10.2\mathrm{eV}$

$13.6\left(\frac{1}{{1}^{2}}-\frac{1}{{3}^{2}}\right)\mathrm{eV}=12.1\mathrm{eV}$

$13.6\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)\mathrm{eV}=1.9\mathrm{eV}$

The following table gives the transition corresponding to the energy of the photon:

Energy of photon | Transition |

12.1 eV | n = 3 to n = 1 |

10.2 eV | n = 2 to n = 1 |

1.9 eV | n = 3 to n = 2 |

A hydrogen atom consists of only one electron. An electron can have transitions, like from

*n*= 3 to

*n*= 2 or from

*n*= 2 to

*n*= 1, at a time.

So, it can be concluded that the photons are emitted either from three atoms (when all the three transitions of electrons are in different atoms) or from two atoms (when an atom has

*n*= 3 to

*n*= 2 and then

*n*= 2 to

*n*= 1 electronic transition and the other has

*n*= 3 to

*n*= 1 electronic transition).

#### Page No 383:

#### Question 3:

Suppose, the electron in a hydrogen atom makes transition from *n* = 3 to *n* = 2 in 10^{−8} s. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is

(a) 10^{−34}^{}N m

(b) 10^{−24}^{}N m

(c) 10^{−42}^{}N m

(d) 10^{−8}^{}N m

#### Answer:

(c) 10^{−42}^{}N-m

The angular momentum of the electron for the *n*th state is given by

${L}_{\mathrm{n}}=\frac{n\mathrm{h}}{2\mathrm{\pi}}$

Angular momentum of the electron for *n* = 3, ${L}_{\mathrm{i}}=\frac{3h}{2\mathrm{\pi}}$

Angular momentum of the electron for *n* = 2, ${L}_{\mathrm{f}}=\frac{2h}{2\mathrm{\pi}}$

The torque is the time rate of change of the angular momentum.

$\mathrm{Torque},\tau =\frac{{L}_{\mathrm{f}}-{L}_{\mathrm{i}}}{t}\phantom{\rule{0ex}{0ex}}=\frac{(2\mathrm{h}/2\mathrm{\pi})-(3\mathrm{h}/2\mathrm{\pi})}{{10}^{-8}}\phantom{\rule{0ex}{0ex}}=\frac{-(\mathrm{h}/2\mathrm{\pi})}{{10}^{-8}}\phantom{\rule{0ex}{0ex}}=\frac{-{10}^{-34}}{{10}^{-8}}\left[\because \frac{\mathrm{h}}{2\mathrm{\pi}}\approx {10}^{-34}\mathrm{J}-\mathrm{s}\right]\phantom{\rule{0ex}{0ex}}=-{10}^{-42}\mathrm{N}-\mathrm{m}\phantom{\rule{0ex}{0ex}}$

The magnitude of the torque is 10^{$-$42 }N-m.

#### Page No 383:

#### Question 4:

In which of the following transitions will the wavelength be minimum?

(a) *n* = 5 to *n* = 4

(b) *n* = 4 to *n* = 3

(c) *n* = 3 to *n* = 2

(d) *n* = 2 to *n* = 1

#### Answer:

(d) *n* = 2 to *n* = 1

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by

$\frac{1}{\lambda}=\mathrm{R}{Z}^{2}\left(\frac{1}{{n}_{1}}-\frac{1}{{n}_{2}}\right)$

Here, R is the Rydberg constant.

For the transition from *n* = 5 to *n* = 4, the wavelength is given by

$\frac{1}{\lambda}=\mathrm{R}{Z}^{2}\left(\frac{1}{{4}^{2}}-\frac{1}{{5}^{2}}\right)\phantom{\rule{0ex}{0ex}}\lambda =\frac{400}{9\mathrm{R}{Z}^{2}}$

For the transition from *n* = 4 to *n* = 3, the wavelength is given by

$\frac{1}{\lambda}=\mathrm{R}{Z}^{2}\left(\frac{1}{{3}^{2}}-\frac{1}{{4}^{2}}\right)\phantom{\rule{0ex}{0ex}}\lambda =\frac{144}{7\mathrm{R}{Z}^{2}}$

For the transition from *n* = 3 to *n* = 2, the wavelength is given by

$\frac{1}{\lambda}=\mathrm{R}{Z}^{2}\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)\phantom{\rule{0ex}{0ex}}\lambda =\frac{36}{5\mathrm{R}{Z}^{2}}$

For the transition from *n* = 2 to *n* = 1, the wavelength is given by

$\frac{1}{\lambda}=\mathrm{R}{Z}^{2}\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}\lambda =\frac{2}{\mathrm{R}{Z}^{2}}$

From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from* n* = 2 to *n* = 1 will be minimum.

#### Page No 383:

#### Question 5:

In which of the following systems will the radius of the first orbit (*n* = 1) be minimum?

(a) Hydrogen atom

(b) Deuterium atom

(c) Singly ionized helium

(d) Doubly ionized lithium

#### Answer:

(d) Doubly ionized lithium

For a hydrogen-like ion with *Z* protons in the nucleus, the radius of the* n*th state is given by

${r}_{\mathrm{n}}=\frac{{n}^{2}{a}_{0}}{Z}$

Here, *a*_{0} = 0.53 pm

For lithium,

Z = 3

Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

#### Page No 383:

#### Question 7:

Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of the principal quantum number *n*?

Figure

#### Answer:

(c)

The speed (*v*) of electron can be expressed as

$v=\frac{Z{e}^{2}}{2{\in}_{0}hn}$ ....(1)

Here,*Z* = Number of protons in the nucleus*e* = Magnitude of charge on electron charge*n* = Principal quantum number*h* = Planck's constant

It can be observed from equation (1) that the velocity of electron is inversely proportional to the principal quantum number (*n*).

Therefore, the graph between them must be a rectangular hyperbola.

The correct curve is (c).

#### Page No 383:

#### Question 8:

As one considers orbits with higher values of *n* in a hydrogen atom, the electric potential energy of the atom

(a) decreases

(b) increases

(c) remains the same

(d) does not increase

#### Answer:

(b) increases

The electric potential energy of hydrogen atom with electron at the* n*th state is given by

$V=-\frac{2\times 13.6}{{n}^{2}}$

As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.

#### Page No 383:

#### Question 9:

The energy of an atom (or ion) in its ground state is −54.4 eV. It may be

(a) hydrogen

(b) deuterium

(c) He^{+}

(d) Li^{++}

#### Answer:

(c) He^{+}

The total energy of a hydrogen-like ion, having *Z* protons in its nucleus, is given by

$E=-\frac{13.6{Z}^{2}}{{n}^{2}}$ eV

Here, *n* = Principal quantum number

For ground state,*n* = 1

∴ Total energy*, E* = $-$ 13.6 *Z*^{2} eV

For hydrogen,*Z* = 1

∴ Total energy*, E *= $-$ 13.6 eV

For deuterium,*Z* = 1

∴ Total energy*, E* = $-$ 13.6 eV

For He^{+},*Z* = 2* *∴ Total energy*, E* = $-$ 13.6×2^{2} = $-$ 54.4 eV

For Li^{++},*Z* = 3

∴ Total energy*, E* = $-$ 13.6×3^{2} = $-$ 122.4 eV

Hence, the ion having an energy of −54.4 eV in its ground state may be He^{+}.

#### Page No 383:

#### Question 10:

The radius of the shortest orbit in a one-electron system is 18 pm. It may be

(a) hydrogen

(b) deuterium

(c) He^{+}

(d) Li^{++}

#### Answer:

(d) Li^{++}

The radius of the *n*th orbit in one electron system is given by

${r}_{\mathrm{n}}=\frac{{n}^{2}{a}_{0}}{Z}$

Here, *a*_{0} = 53 pm

For the shortest orbit,*n* = 1

For hydrogen,*Z* = 1

∴ Radius of the first state of hydrogen atom = 53 pm

For deuterium,

Z= 1

∴ Radius of the first state of deuterium atom = 53 pm

For He^{+},*Z* = 2

∴ Radius of He^{+} atom = $\frac{53}{2}\mathrm{pm}=26.5\mathrm{pm}$

For Li^{++},*Z* = 3

∴ Radius of Li^{++} atom = $\frac{53}{3}\mathrm{pm}=17.66\mathrm{pm}\approx 18\mathrm{pm}$

The given one-electron system having radius of the shortest orbit to be 18 pm may be Li^{++}.

#### Page No 383:

#### Question 11:

A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by

(a) 1.05 × 10^{−34} J s

(b) 2.11 × 10^{−34} J s

(c) 3.16 × 10^{−34} J s

(d) 4.22 × 10^{−34} J s

#### Answer:

(a) 1.05 × 10^{−34} J s

Let after absorption of energy, the hydrogen atom goes to the *n*th excited state.

Therefore, the energy absorbed can be written as

$10.2=13.6\times \left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{10.2}{13.6}=1-\frac{1}{{n}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{n}^{2}}=\frac{13.6-10.2}{13.6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{n}^{2}}=\frac{3.4}{13.6}\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow n=2\phantom{\rule{0ex}{0ex}}$

The orbital angular momentum of the electron in the *n*th state is given by

${L}_{\mathrm{n}}=\frac{nh}{2\pi}$

Change in the angular momentum, $\u2206L=\frac{2h}{2\pi}-\frac{h}{2\pi}=\frac{h}{2\pi}$

∴ $\u2206L=1.05\times {10}^{-34}\mathrm{Js}$

#### Page No 383:

#### Question 12:

Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground states?

(a) Radius of the orbit

(b) Speed of the electron

(c) Energy of the atom

(d) Orbital angular momentum of the electron

#### Answer:

(d) Orbital angular momentum of the electron

According to Bohr's atomic theory, the orbital angular momentum of an electron in a one-electron system is given by

${L}_{\mathrm{n}}=\frac{nh}{2\mathrm{\pi}}$

Here,*n* = Principal quantum number

The angular momentum is independent of the atomic number of the one-electron system. Therefore, it is same for all hydrogen-like atoms and ions in their ground states.

The other parameters given here are dependent on the atomic number of the hydrogen-like atom or ion taken.

#### Page No 383:

#### Question 13:

In a laser tube, all the photons

(a) have same wavelength

(b) have same energy

(c) move in same direction

(d) move with same speed

#### Answer:

(d) move with same speed

All the photons emitted in the laser move with the speed equal to the speed of light (*c* = 3×10^{8} m/s).

Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.

#### Page No 383:

#### Question 1:

In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory

(a) the amount of hydrogen taken is much smaller than that present in the star

(b) the temperature of hydrogen is much smaller than that of the star

(c) the pressure of hydrogen is much smaller than that of the star

(d) the gravitational pull is much smaller than that in the star

#### Answer:

(b) the temperature of hydrogen is much smaller than that of the star

The number of lines of the hydrogen spectrum depends on the excitation of the hydrogen atom. This is dependent on the heat energy absorbed by the hydrogen atoms. More the temperature of the hydrogen sample, more is the heat energy. The temperature of hydrogen at the star is much more than that can be produced in the laboratory. Hence, less number of lines are observed in the hydrogen spectrum in the laboratory than that in a star.

#### Page No 384:

#### Question 2:

An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision

(a) must be elastic

(b) may be partially elastic

(c) must be completely inelastic

(d) may be completely inelastic

#### Answer:

(a) must be elastic.

The minimum energy required to excite a hydrogen atom from its ground state to 1st excited state is approximately 10 eV. As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elestic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat shlould have been produced but it is not so which implies that the collision is completely elastic.

#### Page No 384:

#### Question 3:

Which of the following products in a hydrogen atom are independent of the principal quantum number *n*? The symbols have their usual meanings.

(a) *vn*

(b) E*r*

(c) E*n*

(d) *vr*

#### Answer:

(a) *vn*

(b) E*r*

Relations for energy, radius of the orbit and its velocity are given by

$E=-\frac{\mathrm{m}{Z}^{2}{\mathrm{e}}^{4}}{8{{\in}_{0}}^{2}{\mathrm{h}}^{2}{n}^{2}}\phantom{\rule{0ex}{0ex}}r=\frac{{\in}_{0}{\mathrm{h}}^{2}{n}^{2}}{{\mathrm{\pi mZe}}^{2}}\phantom{\rule{0ex}{0ex}}v=\frac{Z{\mathrm{e}}^{2}}{2{\in}_{0}\mathrm{h}n}$

Where*Z* : the atomic number of hydrogen like atom

e : electric charge

h : plank constant

m : mass of electron*n* : principal quantam number of the electron

${\in}_{0}$ : permittivity of vacuum

From these relations, we can see that the products independent of *n* are *vn*, E*r*.

#### Page No 384:

#### Question 4:

Let A_{n} be the area enclosed by the *n*th orbit in a hydrogen atom. The graph of ln (A_{n}/A_{1}) against ln(*n*)

(a) will pass through the origin

(b) will be a straight line with slope 4

(c) will be a monotonically increasing nonlinear curve

(d) will be a circle

#### Answer:

(a) will pass through the origin

(b) will be a straight line with slope 4

The radius of the *n*th orbit of a hydrogen atom is given by

${r}_{\mathrm{n}}={n}^{2}{a}_{0}$

Area of the nth orbit is given by

${A}_{\mathrm{n}}=\mathrm{\pi}{r}_{\mathrm{n}}^{2}=\mathrm{\pi}{n}^{4}{a}_{0}^{2}\phantom{\rule{0ex}{0ex}}{A}_{1}=\mathrm{\pi}{a}_{0}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ln}\left(\frac{{A}_{\mathrm{n}}}{{A}_{1}}\right)=\mathrm{ln}\left(\frac{\mathrm{\pi}{n}^{4}{a}_{0}^{2}}{\mathrm{\pi}{a}_{0}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(\frac{{A}_{\mathrm{n}}}{{A}_{1}}\right)=4\mathrm{ln}n...\left(1\right)$

From the above expression, the graph of ln (A_{n}/A_{1}) against ln(*n*) will be a straight line passing through the origin and having slope 4.

#### Page No 384:

#### Question 5:

Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let *r*, *u*, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state

(a) *r*_{A} > *r*_{B}

(b) *u*_{A} > *u*_{B}

(c) E_{A} > E_{B}

(d) L_{A} > L_{B}

#### Answer:

(b) *u*_{A} > *u*_{B}

The ionisation energy of a hydrogen like ion of atomic number *Z* is given by

$V=(13.6\mathrm{eV})\times {Z}^{2}$

Thus, the atomic number of ion A is greater than that of B (*Z*_{A} > *Z*_{B}).

The radius of the orbit is inversely proportional to the atomic number of the ion.

∴ *r*_{A} > *r*_{B}

Thus, (a) is incorrect.

The speed of electron is directly proportional to the atomic number.

Therefore, the speed of the electron in the orbit of A will be more than that in B.

Thus, *u*_{A} > *u*_{B}_{}is correct.

The total energy of the atom is given by

$E=-\frac{m{Z}^{2}{e}^{2}}{8{\in}_{0}{h}^{2}{n}^{2}}$

As the energy is directly proportional to *Z*^{2}, the energy of A will be less than that of B, i.e. *E*_{A} < *E*_{B}.

The orbital angular momentum of the electron is independent of the atomic number.

Therefore, the relation *L*_{A} > *L*_{B }is invalid.

#### Page No 384:

#### Question 6:

When a photon stimulates the emission of another photon, the two photons have

(a) same energy

(b) same direction

(c) same phase

(d) same wavelength

#### Answer:

(a) same energy

(b) same direction

(c) same phase

(d) same wavelength

When a photon stimulates the emission of another photon, the two photons have same energy, direction, phase, and wavelength or we can say that the two photons are coherent.

When an atom is present in its excited state then if a photon of energy equal to the energy gap between the excited state and any lower stable state is incident on this atom then the atom transits from upper state to the lower stable state by emitting a photon of energy equal to the energy gap between the two states. It is called stimulated emission. The emitted photon and incident photon have same energy and hence same wavelength. Also these two photons will be in phase and in the same direction. This process of producing monochromatic and unidirectional light is called lasing action.

#### Page No 384:

#### Question 1:

The Bohr radius is given by ${a}_{0}=\frac{{\mathrm{\epsilon}}_{0}{h}^{2}}{\mathrm{\pi}m{e}^{2}}$. Verify that the RHS has dimensions of length.

#### Answer:

The dimensions of *ε*_{0} can be derived from the formula given below:

$a=\frac{{\epsilon}_{0}{h}^{2}}{\mathrm{\pi}m{e}^{2}}=\frac{{\mathrm{A}}^{2}{\mathrm{T}}^{2}{\left({\mathrm{ML}}^{2}{\mathrm{T}}^{-1}\right)}^{2}}{{\mathrm{L}}^{2}{\mathrm{ML}}^{-2}\mathrm{M}{\left(\mathrm{AT}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{M}}^{2}{\mathrm{L}}^{2}{\mathrm{T}}^{-2}}{{\mathrm{M}}^{2}{\mathrm{L}}^{3}{\mathrm{T}}^{-2}}=\mathrm{L}$

Clearly, *a*_{0} has the dimensions of length.

#### Page No 384:

#### Question 2:

Find the wavelength of the radiation emitted by hydrogen in the transitions (a) *n* = 3 to *n* = 2, (b) *n* = 5 to *n* = 4 and (c) *n* = 10 to *n* = 9.

#### Answer:

From Balmer empirical formula, the wavelength $\left(\lambda \right)$ of the radiation is given by

$\frac{1}{\lambda}=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

Here,* R *= Rydberg constant = 1.097$\times $10^{7 }m^{$-1$}

* **n*_{1} = Quantum number of final state

*n*_{2} = Quantum number of initial state

(a)

For transition from *n* = 3 to *n* = 2:

Here*,**n*_{1} = 2*n*_{2} = 3

$\frac{1}{\mathrm{\lambda}}=1.09737\times {10}^{7}\times \left(\frac{1}{4}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=\frac{36}{5\times 1.09737\times {10}^{7}}\phantom{\rule{0ex}{0ex}}=6.56\times {10}^{-7}=656\mathrm{nm}$

(b)

For transition from *n* = 5 to *n* = 4:

Here,*n*_{1} = 4*n*_{2} = 5

$\frac{1}{\lambda}=1.09737\times {10}^{-7}\left(\frac{1}{16}-\frac{1}{25}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{400}{1.09737\times {10}^{7}\times 9}\phantom{\rule{0ex}{0ex}}=4050\mathrm{nm}$

(c)

For transition from *n* = 10 to *n* = 9:

Here,*n*_{1} = 9*n*_{2} = 10

$\frac{1}{\lambda}=1.09737\times {10}^{7}\left(\frac{1}{81}-\frac{1}{100}\right)\phantom{\rule{0ex}{0ex}}\lambda =\frac{81\times 100}{19\times 1.09737\times {10}^{7}}\phantom{\rule{0ex}{0ex}}=38849\mathrm{nm}$

#### Page No 384:

#### Question 3:

Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He^{+} and (c) Li^{++}.

#### Answer:

Given:

For the smallest wavelength, energy should be maximum.

Thus, for maximum energy, transition should be from infinity to the ground state.

∴ *n*_{1}_{}= 1 *n*_{2}= $\infty $

(a) Wavelength of the radiation emitted $\left(\lambda \right)$ is given by

$\frac{\mathit{1}}{\lambda}=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

For hydrogen,

Atomic number, *Z* = 1*R* = Rydberg constant = 1.097×10^{7} m^{$-$1}

On substituting the respective values,

$\lambda =\frac{1}{1.097\times {10}^{7}}=\frac{1}{1.097}\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=0.911\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=91.1\times {10}^{-9}=91\mathrm{nm}$

(b)

For He^{+},

Atomic number, *Z* = 2

Wavelength of the radiation emitted by He^{+ }($\lambda $) is given by

$\frac{\mathit{1}}{\lambda}=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

$\therefore \frac{1}{\lambda}={\left(2\right)}^{2}(1.097\times {10}^{7})\left(\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(\infty \right)}^{2}}\right)$

$\Rightarrow \lambda =\frac{91\mathrm{nm}}{4}=23\mathrm{nm}$

(c) For Li^{++},

Atomic number, *Z* = 3

Wavelength of the radiation emitted by Li^{++ }($\lambda $) is given by

$\frac{\mathit{1}}{\lambda}=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

$\therefore \frac{1}{\lambda}={\left(3\right)}^{2}\times (1.097\times {10}^{7})\left(\frac{1}{{1}^{2}}-\frac{1}{{\infty}^{2}}\right)$

$\Rightarrow \lambda =\frac{91\mathrm{nm}}{{Z}^{2}}=\frac{91}{9}=10\mathrm{nm}$

#### Page No 384:

#### Question 4:

Evaluate Rydberg constant by putting the values of the fundamental constants in its expression.

#### Answer:

Expression of Rydberg constant (*R*) is given by

$R=\frac{m{e}^{4}}{8{h}^{2}c{\in}_{0}^{2}}$

Mass of electron, *m*_{e} = 9.31$\times $10^{21} kg

Charge, *e* = 1.6 × 10^{−19} C

Planck's constant,* h* = 6.63 × 10^{−34} J-s,

Speed of light, *c* = 3 × 10^{8} m/s,

Permittivity of vacuum, ∈_{0} = 8.85 × 10^{−12}^{ }C^{2}N^{$-1$}m

On substituting the values in the expression, we get^{$R=\frac{\left(9.31\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-19}\right)}^{4}}{8\times {\left(6.63\times {10}^{-34}\right)}^{2}\times \left(3\times {10}^{8}\right)\times {\left(8.85\times {10}^{-12}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow R=1.097\times {10}^{7}{\mathrm{m}}^{-1}$}

#### Page No 384:

#### Question 5:

Find the binding energy of a hydrogen atom in the state *n* = 2.

#### Answer:

The binding energy (E) of hydrogen atom is given by

$E=\frac{-13.6}{{n}^{2}}\mathrm{eV}$

For state n = 2,

$E=-\frac{13.6}{{\left(2\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=-3.4\mathrm{eV}$

Thus, binding energy of hydrogen at *n* = 2 is $-3.4\mathrm{eV}$.

#### Page No 384:

#### Question 6:

Find the radius and energy of a He^{+} ion in the states (a) *n* = 1, (b) *n* = 4 and (c) *n* = 10.

#### Answer:

For He^{+} ion,

Atomic number, *Z *= 2

For hydrogen like ions, radius $\left(r\right)$ of the nth state is given by

$r=\frac{0.53{n}^{2}}{Z}$Å

Here,*Z* = Atomic number of ions*n* = Quantum number of the state

Energy $\left(E\right)$ of the nth state is given by*E*_{n} = $-\frac{13.6{Z}^{2}}{{n}^{2}}$

(a)

For *n* = 1,

Radius, $r=\frac{0.53\times {\left(1\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}=0.265{\mathrm{A}}^{0}$ Å

Energy, *E*_{n} = $\frac{-13.6\times 4}{1}$

= $-$ 54.4 eV

(b)

For *n* = 4,

Radius, $r=\frac{0.53\times 16}{2}=4.24\mathrm{\AA}$

Energy, $E=\frac{-13.6\times 4}{16}=-3.4e\mathrm{V}$

(c)

For *n *= 10,

Radius, $r=\frac{0.53\times 100}{2}$

= 26.5 Å

Energy, $E=\frac{-13.6\times 4}{100}=-0.544\mathrm{eV}$

#### Page No 384:

#### Question 7:

A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?

#### Answer:

Wavelength of ultraviolet radiation, $\lambda $ = 102.5 nm = 102,5 $\times $10^{$-9$} m

Rydberg's constant, *R* = 1.097$\times $ 10^{7} m^{$-1$}

Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.

Lyman series is obtained when an electron jumps to the ground state (*n*_{â€‹1 }= 1) from any excited state (*n*_{â€‹2}).

Wavelength of light $\left(\lambda \right)$ is given by

$\frac{1}{\lambda}=R\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Here},R=\mathrm{Rydberg}\mathrm{constant}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{102.5\times {10}^{-9}}=1.097\times {10}^{7}\left(\frac{1}{{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{10}^{9}}{102.5}=1.097\times {10}^{7}\left(1-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{10}^{2}}{102.5}=1.097\left(1-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1-\frac{1}{{n}_{2}^{2}}=\frac{100}{102.5\times 1.097}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{n}_{2}^{2}}=1-\frac{100}{102.5\times 1.097}\phantom{\rule{0ex}{0ex}}\Rightarrow {n}_{2}=\sqrt{9.041}=3$

The transition will be from 1 to 3.

#### Page No 384:

#### Question 8:

(a) Find the first excitation potential of He^{+} ion. (b) Find the ionization potential of Li^{++} ion.

#### Answer:

(a) PE of hydrogen like atom in the *n*th state, V = $=\frac{-13.6{Z}^{2}}{{n}^{2}}\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

Here, *Z* is the atomic number of that atom.

For the first excitation, the atom has to be excited from n = 1 to n = 2 state.

So, its excitation potential will be equal to the difference in the potential of the atom in n = 1 and in n = 2 states.

First excitation potential of He^{+}

$-13.6{Z}^{2}(1-\frac{1}{{2}^{2}})eV\phantom{\rule{0ex}{0ex}}=-10.2{Z}^{2}eV$

⇒10.2 × Z^{2}

= 10.2 × 4

= 40.8 V

(b) Ionization Potential Li^{++} = 13.6 V × Z^{2}

= 13.6 × 9

= 122.4 V

#### Page No 384:

#### Question 9:

A group of hydrogen atoms are prepared in *n* = 4 states. List the wavelength that are emitted as the atoms make transitions and return to *n* = 2 states.

#### Answer:

There will be three wavelengths.

(i) For the transition from (*n* = 4) to (*n *= 3) state

(ii) For the transition from (*n* = 3) to (*n *= 2) state

(iii) For the transition from (*n* = 4) to (*n *= 2) state

Let ($\lambda $_{1}) be the wavelength when the atom makes transition from (*n* = 4) state to (*n* = 2) state._{â€‹ }

Here,*n*_{1}_{ }= 2*n*_{2}_{ }= 4

Now, the wavelength ($\lambda $_{1})â€‹ will be

$\frac{1}{{\lambda}_{1}}=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$R=1.097\times {10}^{7}{\mathrm{m}}^{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda}_{1}}=1.097\times {10}^{7}\times \left(\frac{1}{4}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{\lambda}_{1}}=1.097\times {10}^{7}\left(\frac{4-1}{16}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{\lambda}_{1}}=\frac{1.097\times {10}^{7}\times 3}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{1}=\frac{16\times {10}^{-7}}{3\times 1.097}\phantom{\rule{0ex}{0ex}}=4.8617\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=486.1\times {10}^{-9}\phantom{\rule{0ex}{0ex}}=487\mathrm{nm}$

When an atom makes transition from (*n* = 4) to (*n *= 3), the wavelength ($\lambda $2) is given by

$\mathrm{Here}\mathrm{again},\phantom{\rule{0ex}{0ex}}{n}_{1}=3\phantom{\rule{0ex}{0ex}}{n}_{2}=4\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda}_{2}}=1.097\times {10}^{7}\left(\frac{1}{9}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{\lambda}_{2}}=1.097\times {10}^{7}\left(\frac{16-9}{144}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{\lambda}_{2}}=\frac{1.097\times {10}^{7}\times 7}{144}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{2}=\frac{144}{1.097\times {10}^{7}\times 7}\phantom{\rule{0ex}{0ex}}=1875\mathrm{nm}$

Similarly, wavelength ($\lambda $_{3}) for the transition from (*n* = 3) to (*n *= 2) is given by

When the transition is *n*_{1} = 2 to *n*_{2} = 3:

$\frac{1}{{\lambda}_{3}}=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{\lambda}_{3}}=1.097\times {10}^{7}\left(\frac{9-4}{36}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{\lambda}_{3}}=\frac{1.097\times {10}^{7}\times 5}{36}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{3}=\frac{36\times {10}^{-7}}{1.097\times 5}=656\mathrm{nm}$

#### Page No 384:

#### Question 10:

A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion.

#### Answer:

Given:

Wavelength of photon,* λ* = 228 Å

Energy $\left(E\right)$ is given by

$E=\frac{h\mathrm{c}}{\lambda}\phantom{\rule{0ex}{0ex}}$

Here*, *c = Speed of light * h *= Planck's constant

$\therefore E=\frac{\left(6.63\times {10}^{-34}\right)\times \left(3\times {10}^{8}\right)}{\left(228\times {10}^{-10}\right)}\phantom{\rule{0ex}{0ex}}=0.0872\times {10}^{-16}\mathrm{J}$

As the transition takes place from *n* = 1 to *n* = 2, the excitation energy (*E*_{1}) will be

${E}_{1}=Rh\mathrm{c}{Z}^{2}\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\phantom{\rule{0ex}{0ex}}{E}_{1}=\left(13.6\mathrm{eV}\right)\times {Z}^{2}\times \left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{1}=\left(13.6\mathrm{eV}\right)\times {Z}^{2}\times \frac{3}{4}$

This excitation energy should be equal to the energy of the photon.

$\therefore 13.6\times \frac{3}{4}\times {\mathrm{Z}}^{2}=0.0872\times {10}^{-16}\phantom{\rule{0ex}{0ex}}{Z}^{2}=\frac{0.0872\times {10}^{-16}\times 4}{13.6\times 3\times 1.6\times {10}^{-19}}=5.34\phantom{\rule{0ex}{0ex}}Z=\sqrt{5.34}=2.3$

The ion may be helium.

#### Page No 384:

#### Question 11:

Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.

#### Answer:

Charge on the electron, *q*_{1} = 1.6$\times $10 ^{$-19$ }C

Charge on the nucleus, *q*_{2} = 1.6$\times $10 ^{$-19$ }C

Let *r *be the distance between the nucleus and the electron.

Coulomb force $\left(F\right)$ is given by

$F=\frac{{q}_{1}{q}_{2}}{4\mathrm{\pi}{\in}_{0}{r}^{2}}$ .....(1)

Here*, **q*_{1} = *q*_{2} =* q* = 1.6$\times $10 ^{$-19$ }C

Smallest distance between the nucleus and the first orbit, *r* = 0.53$\times $10^{$-$10} m

$K=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}$ = 9$\times $10^{9} Nm^{2}C^{$-$2}

Substituting the respective values in (1), we get

$F=\frac{\left(9\times {10}^{9}\right)\times \left(1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{{\left(0.53\times {10}^{-10}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times 1.6\times 9\times {10}^{-9}}{{\left(0.53\right)}^{2}}=82.02\times {10}^{-9}\phantom{\rule{0ex}{0ex}}=8.2\times {10}^{-8}\mathrm{N}$

#### Page No 384:

#### Question 12:

A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers *n* of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.

#### Answer:

(a)

The binding energy of hydrogen is given by

$E=\frac{13.6}{{n}^{2}}\mathrm{eV}$

For binding energy of 0.85 eV,

${{n}_{2}}^{2}=\frac{13.6}{0.85}=16\phantom{\rule{0ex}{0ex}}{n}_{2}=4$

For binding energy of 10.2 eV,

${{n}_{1}}^{2}=\frac{13.6}{10.2}\phantom{\rule{0ex}{0ex}}{n}_{1}=1.15\phantom{\rule{0ex}{0ex}}\Rightarrow {n}_{1}=2\phantom{\rule{0ex}{0ex}}$

The quantum number of the upper and the lower energy state are 4 and 2, respectively.

(b) Wavelength of the emitted radiation $\left(\lambda \right)$ is given by

$\frac{1}{\lambda}=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

Here,

R = Rydberg constant *n*_{1} and* **n*_{2} are quantum numbers.

$\therefore \frac{1}{\lambda}=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=\frac{16}{1.097\times 3\times {10}^{7}}\phantom{\rule{0ex}{0ex}}=4.8617\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=487\mathrm{nm}$

#### Page No 384:

#### Question 13:

Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?

#### Answer:

As the second photon emitted lies in the Lyman series, the transition will be from the states having quantum numbers *n* = 2 to *n* = 1.

Wavelength of radiation $\left(\lambda \right)$ is given by

$\frac{1}{\lambda}=R\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}$

Here, R is the Rydberg constant, having the value of 1.097×10^{7} m^{-1}.

$\frac{1}{\lambda}=1.097\times {10}^{7}\left[\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(2\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda}=1.097\times {10}^{7}\left[1-\frac{1}{4}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda}=1.097\times \frac{3}{4}\times {10}^{7}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{4}{1.097\times 3\times {10}^{7}}\phantom{\rule{0ex}{0ex}}=1.215\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=121.5\times {10}^{-9}=122\mathrm{nm}$

#### Page No 384:

#### Question 14:

A hydrogen atom in state *n* = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of *n* in the intermediate state?

#### Answer:

Energy (*E*) of the* n*th state of hydrogen atom is given by*E *= $\frac{13.6}{{n}^{2}}$ eV

For* n* = 6,

$\therefore E=\frac{-13.6}{36}=-0.377777777$ eV

Energy of hydrogen atom in the ground state = −13.6 eV

Energy emitted in the second transition = Energy of ground state $-$ (Energy of hydrogen atom in the 6th state + Energy of photon)

$=13.6-\left(0.3777\overline{)7}+1.13\right)\phantom{\rule{0ex}{0ex}}=12.09=12.1\mathrm{eV}$

(b)

Energy in the intermediate state = (Energy of photon emitted in the first transition) + (Energy of n = 6 state)

= 1.13 eV + 0.377 eV

= 1.507 eV

Energy of the *n*th state can expressed as

$\frac{13.6}{{n}^{2}}=1.507\phantom{\rule{0ex}{0ex}}$

$\therefore n=\sqrt{\frac{13.6}{1.507}}=\sqrt{9.024}\approx 3$

#### Page No 384:

#### Question 15:

What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?

#### Answer:

In ground state, the potential energy of a hydrogen atom is zero.

An electron is bound to the nucleus with an energy of 13.6 eV.

Therefore, we have to give 13.6 eV energy to move the electron from the nucleus.

Let us calculate the excitation energy required to take an atom from the ground state (*n* = 1) to the first excited state (*n* = 2).

$E=13.6\times \left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\mathrm{eV}$

Therefore, the excitation energy is given by

$E=13.6\times \left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\mathrm{eV}\phantom{\rule{0ex}{0ex}}E=13.6\times \frac{3}{4}\mathrm{eV}=10.2\mathrm{eV}$

Energy of 10.2 eV is needed to take an atom from the ground state to the first excited state.

∴ Total energy of an atom in the first excitation state = 13.6 eV + 10.2 eV = 23.8 eV

#### Page No 384:

#### Question 16:

A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.

#### Answer:

Given:

Energy (E) of the ground state will be the energy acquired in the transition of the 2 excitation state to ground state.

${E}_{1}=\frac{hc}{{\mathrm{\lambda}}_{1}}\phantom{\rule{0ex}{0ex}}$

Here,*h *= Planck's constant*c* = Speed of light

${\lambda}_{1}$ = Wavelength of the radiation emitted when atoms come from the highest excited state to ground state

$\therefore {E}_{1}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(46\times {10}^{-9})}\mathrm{J}\phantom{\rule{0ex}{0ex}}{E}_{1}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(46\times {10}^{-9})\times (1.6\times {10}^{-19})}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=\frac{1242}{46}=27\mathrm{eV}$

Energy in the first excitation state $\left({E}_{2}\right)$ will be the energy acquired in the transition of the highest energy state to the 2nd excitation state.

${E}_{2}=\frac{hc}{{\mathrm{\lambda}}_{n}}\phantom{\rule{0ex}{0ex}}$

Here,

${\lambda}_{n}$ = Wavelength of the radiation emitted when an atom comes from the highest energy state to the 2nd excitation stateâ€‹.

${E}_{2}=\frac{hc}{{\mathrm{\lambda}}_{n}}\phantom{\rule{0ex}{0ex}}{E}_{2}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(103.5\times {10}^{-9})}\mathrm{J}\phantom{\rule{0ex}{0ex}}{E}_{2}=\frac{(6.63\times {10}^{-34})\times (3\times {10}^{8})}{(103.5\times {10}^{-9})\times (1.6\times {10}^{-19})}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=12\mathrm{eV}$

#### Page No 384:

#### Question 17:

A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.

#### Answer:

(a) If the atom is excited to the principal quantum* (n), *then the number of transitions is given by

$\frac{n\left(n-1\right)}{2}\phantom{\rule{0ex}{0ex}}$

It is given that a total of 6 photons are emitted. Therefore, total number of transitions is 6.

∴ $\frac{n\left(n-1\right)}{2}\phantom{\rule{0ex}{0ex}}$ = 6

⇒ *n* = 4

Thus, the principal quantum number is 4 and the gas is in the 4th excited state.

#### Page No 384:

#### Question 18:

Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.

#### Answer:

Let the mass of the electron be* m. *

Let the radius of the hydrogen's first stationary orbit be *r.*

Let the linear speed and the angular speed of the electron be *v* and *ω,* respectively.

According to the Bohr's theory, angular momentum (*L*) of the electron is an integral multiple of *h*/2$\mathrm{\pi}$, where* h* is the Planck's constant.

$\Rightarrow mvr=\frac{nh}{2\mathrm{\pi}}$ (Here, *n* is an integer.)

$v=r\omega \phantom{\rule{0ex}{0ex}}\Rightarrow m{r}^{2}\mathrm{\omega}=\frac{nh}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =\frac{nh}{2\mathrm{\pi}\times m\times {r}^{2}}\phantom{\rule{0ex}{0ex}}$

$\therefore \omega =\frac{1\times \left(6.63\times {10}^{-34}\right)}{2\times \left(3.14\right)\times \left(9.1093\times {10}^{-31}\right)\times {\left(0.53\times {10}^{-10}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.413\times {10}^{17}\mathrm{rad}/\mathrm{s}=4.13\times {10}^{16}\mathrm{rad}/\mathrm{s}$

#### Page No 384:

#### Question 19:

A spectroscopic instrument can resolve two nearby wavelengths λ and λ + Δλ if λ/Δλ is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?

#### Answer:

The range of wavelength falling in Balmer series is between 656.3 nm and 365 nm.

It is given that the resolution of the instrument is *λ*/ (*λ* + âˆ†*λ*) < 8000.

Number of wavelengths in this range will be calculated in the following way:

$\frac{656.3-365}{8000}=36$

Two lines will be extra for the first and last wavelength.

∴ Total number of lines = 36 + 2 = 38

#### Page No 384:

#### Question 20:

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number *n* changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).

#### Answer:

Given:

Possible transitions:

From *n*_{1} = 1 to *n*_{2} = 3

*n*_{1} = 2 to *n*_{2} = 4

(a) Here, *n*_{1} = 1 and *n*_{2} = 3

Energy, $E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

$E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{8}{9}.....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Energy}\left(E\right)\mathrm{is}\mathrm{also}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}E=\frac{hc}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}$

Here, h = Planck constant

c = Speed of the light

$\lambda $ = Wavelength of the radiation

$\phantom{\rule{0ex}{0ex}}\therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{\lambda}.....\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Equating}\mathrm{equation}s\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}\times 9}{13.6\times 8}\phantom{\rule{0ex}{0ex}}=0.027\times {10}^{-7}=103\mathrm{nm}$

(b) Visible radiation comes in Balmer series.

As '*n*' changes by 2, we consider *n* = 2 to *n* = 4.

Energy, ${E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

$=13.6\times \left(\frac{1}{4}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}=2.55\mathrm{eV}$

If ${\lambda}_{1}$ is the wavelength of the radiation, when transition takes place between quantum number *n* = 2 to *n* = 4, then

255 = $\frac{1242}{{\lambda}_{1}}$

or *λ*_{1}_{â€‹} = 487 nm

#### Page No 384:

#### Question 21:

According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?

#### Answer:

Let *v*_{0} be the velocity of the electron moving in the ground state and ${r}_{0}$ be the radius of the ground state.

Frequency of the revolution of electron in the circle is given by

$f=\frac{{v}_{0}}{2\mathrm{\pi}{r}_{0}}$

Frequency of the radiation emitted = Frequency of the revolution of electron

∴ Frequency of the radiation emitted = $\frac{{v}_{0}}{2\mathrm{\pi}{r}_{0}}$

Also, *c* = *f*$\lambda $

Here, *c* = Speed of light

$\lambda $ = Wavelength of the radiation emitted

$\Rightarrow $$\lambda =\frac{c}{f}$

$\therefore \lambda =\frac{2\mathrm{\pi}{r}_{0}c}{{v}_{0}}\phantom{\rule{0ex}{0ex}}=\frac{2\times \left(3.14\right)\times \left(53\times {10}^{-12}\right)\times (3\times {10}^{8})}{\left(2.187\times {10}^{6}\right)}\phantom{\rule{0ex}{0ex}}=45.686\times {10}^{-12}\mathrm{m}=45.7\mathrm{nm}$

#### Page No 385:

#### Question 22:

The average kinetic energy of molecules in a gas at temperature T is 1.5 *k*T. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take *k* = 8.62 × 10^{−5} eV K^{−1}.

#### Answer:

Average kinetic energy $\left(K\right)$ of the molecules in a gas at temperature $\left(T\right)$ is given by

K = $\frac{3}{2}kT$

Here,*k* = 8.62 × 10^{−5} eVK^{−1}*T* = Temperature of gas

The binding energy of hydrogen atom is 13.6 eV.

According to the question,

Average kinetic energy of hydrogen molecules = Binding energy of hydrogen atom

$\therefore $1.5 *kT* = 13.6

$\Rightarrow $1.5 × 8.62 × 10^{−5} × *T *= 13.6

$\Rightarrow T=\frac{13.6}{1.5\times 8.62\times {10}^{-5}}\phantom{\rule{0ex}{0ex}}=1.05\times {10}^{5}\mathrm{K}$

No, it is impossible for hydrogen to remain in molecular state at such a high temperature.

#### Page No 385:

#### Question 23:

Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to *n* = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

#### Answer:

Given:

Wavelength of red light, $\lambda $ = 653.1 nm = 653.1$\times $10^{$-9$} m

Kinetic energy of H_{2} molecules $\left(K\right)$ is given by

$K=\frac{3}{2}kT$ ...(1)

Here*, k* = 8.62 × 10^{−5} eV/K

*T* = Temperature of H_{2} molecules

Energy released $\left(E\right)$ when atom goes from ground state to *n* = 3 is given by

$E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

For ground state, *n*_{1} = 1

Also, *n*_{2} = 3

$\therefore E=13.6\left(\frac{1}{1}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=13.6\left(\frac{8}{9}\right)...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Kinetic energy of H_{2} molecules = Energy released when hydrogen atom goes from ground state to *n* = 3 state

$\therefore \frac{3}{2}\times 8.62\times {10}^{-5}\times T=\frac{13.6\times 8}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{13.6\times 8\times 2}{9\times 3\times 8.62\times {10}^{-5}}\phantom{\rule{0ex}{0ex}}=9.4\times {10}^{4}\mathrm{K}$

#### Page No 385:

#### Question 24:

Average lifetime of a hydrogen atom excited to *n* = 2 state is 10^{−8} s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.

#### Answer:

Frequency of electron (*f) *is given by

$f=\frac{m{e}^{4}}{4{{\in}_{0}}^{2}{n}^{3}{h}^{3}}\phantom{\rule{0ex}{0ex}}$

Time period is given by

$T=\frac{1}{f}\phantom{\rule{0ex}{0ex}}T=\frac{4{\in}_{0}^{2}{n}^{3}{h}^{3}}{m{e}^{4}}$

Here,*h* = Planck's constant*m* = Mass of the electron*e* = Charge on the electron

${\epsilon}_{0}$= Permittivity of free space

$\therefore T=\frac{4\times \left(8.85\times {10}^{-12}\right)\times {\left(2\right)}^{3}\times {\left(6.63\times {10}^{-34}\right)}^{3}}{\left(9.10\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-16}\right)}^{4}}\phantom{\rule{0ex}{0ex}}=12247.735\times {10}^{-19}\mathrm{s}$

Average life time of hydrogen, *t* = 10^{$-$8} s

Number of revolutions is given by

$N=\frac{t}{T}\phantom{\rule{0ex}{0ex}}\Rightarrow N=\frac{{10}^{-8}}{12247.735\times {10}^{-19}}$*N* = 8.2 $\times $ 10^{5}^{ }revolution

#### Page No 385:

#### Question 25:

Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.

#### Answer:

Mass of the electron, * m* = 9.1×10^{$-$31}kg

Radius of the ground state,* r = *0.53×10^{$-$10}m

Let * f* be the frequency of revolution of the electron moving in ground state and *A* be the area of orbit.

Dipole moment of the electron (*μ*) is given by*μ* = *ni*A = *qf*A

$=e\times \frac{m{e}^{4}}{4{\in}_{0}^{2}{h}^{3}{n}^{3}}\times \left(\mathrm{\pi}{r}^{2}{n}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{m{e}^{5}\times \left(\mathrm{\pi}{r}^{2}{n}^{2}\right)}{4{\in}_{0}^{2}{h}^{3}{n}^{3}}\phantom{\rule{0ex}{0ex}}$

Here, *h* = Planck's constant*e *= Charge on the electron

${\epsilon}_{0}$ = Permittivity of free space*n* = Principal quantum number

$\therefore \mu =\frac{\left(9.1\times {10}^{-31}\right){\left(1.6\times {10}^{-19}\right)}^{5}\times \mathrm{\pi}\times {\left(0.53\times {10}^{-10}\right)}^{2}}{4\times {\left(8.85\times {10}^{-12}\right)}^{2}\times {\left(6.64\times {10}^{-34}\right)}^{3}\times {\left(1\right)}^{3}}\phantom{\rule{0ex}{0ex}}=0.000917\times {10}^{-20}\phantom{\rule{0ex}{0ex}}=9.176\times {10}^{-24}\mathrm{A}-{\mathrm{m}}^{2}$

#### Page No 385:

#### Question 26:

Show that the ratio of the magnetic dipole moment to the angular momentum (*l* = *mvr*) is a universal constant for hydrogen-like atoms and ions. Find its value.

#### Answer:

Mass of the electron, * m* = 9.1×10^{$-$31}kg

Radius of the ground state,* r = *0.53×10^{$-$10}m

Let * f* be the frequency of revolution of the electron moving in the ground state and *A* be the area of orbit.

Dipole moment of the hydrogen like elements (*μ*) is given by*μ* = *ni*A = *qf*A

$=e\times \frac{m{e}^{4}}{4{\in}_{0}^{2}{h}^{3}{n}^{3}}\times \left(\mathrm{\pi}{r}_{0}^{2}{n}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{m{e}^{5}\times \left(\mathrm{\pi}{{r}_{0}}^{2}{n}^{2}\right)}{4{\in}_{0}^{2}{h}^{3}{n}^{3}}\phantom{\rule{0ex}{0ex}}$

Here, *h* = Planck's constant*e *= Charge on the electron

${\epsilon}_{0}$ = Permittivity of free space*n* = Principal quantum number

Angular momentum of the electron in the hydrogen like atoms and ions (*L)* is given by

$L=mvr=\frac{nh}{2\mathrm{\pi}}$

Ratio of the dipole moment and the angular momentum is given by

$\frac{\mu}{L}=\frac{{e}^{5}\times m\times \mathrm{\pi}{r}^{2}{n}^{2}}{4{\in}_{0}{h}^{3}{n}^{3}}\times \frac{2\mathrm{\pi}}{nh}$

$\frac{\mu}{L}=\frac{{\left(1.6\times {10}^{-19}\right)}^{5}\times \left(9.10\times {10}^{-31}\right)\times {\left(3.14\right)}^{2}\times {\left(0.53\times {10}^{-10}\right)}^{2}}{2\times {\left(8.85\times {10}^{-12}\right)}^{2}\times {\left(6.63\times {10}^{-34}\right)}^{3}\times {1}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\mu}{L}=3.73\times {10}^{10}\mathrm{C}/\mathrm{kg}$

Ratio of the magnetic dipole moment and the angular momentum do not depends on the atomic number '*Z*'.

Hence, it is a universal constant.

#### Page No 385:

#### Question 27:

A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?

#### Answer:

Given:

Minimum wavelength of the light component present in the beam, ${\lambda}_{1}$ = 450 nm

Energy associated $\left({E}_{1}\right)$ with wavelength $\left({\lambda}_{1}\right)$ is given by*E*_{1} = $\frac{hc}{{\lambda}_{1}}$

Here,*c* = Speed of light*h* = Planck's constant

$\therefore {E}_{1}=\frac{1242}{450}$

= 2.76 eV

Maximum wavelength of the light component present in the beam, ${\lambda}_{2}$ = 550 nm

Energy associated $\left({E}_{2}\right)$ with wavelength $\left({\lambda}_{2}\right)$ is given by*E*_{2} = $\frac{hc}{{\lambda}_{2}}$

$\therefore {E}_{2}=\frac{1242}{550}=2.228=2.26\mathrm{eV}$

The given range of wavelengths lies in the visible range.

∴ *n*_{1} = 2, *n*_{2} = 3, 4, 5 ...

Let *E'*_{2} ,* **E'*_{3} , *E'*_{4} and *E'*_{5} be the energies of the 2nd, 3rd, 4th and 5th states, respectively.

$E{\text{'}}_{2}-E{\text{'}}_{3}=13.6\left(\frac{1}{4}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=\frac{12.6\times 5}{30}=1.9\mathrm{eV}\phantom{\rule{0ex}{0ex}}E{\text{'}}_{2}-E{\text{'}}_{4}=13.6\left(\frac{1}{4}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}=2.55\mathrm{eV}\phantom{\rule{0ex}{0ex}}E{\text{'}}_{2}-E{\text{'}}_{5}=13.6\left(\frac{1}{4}-\frac{1}{25}\right)\phantom{\rule{0ex}{0ex}}=\frac{10.5\times 21}{100}=2.856\mathrm{eV}$

Only, *E'*_{2} − *E'*_{4} comes in the range of the energy provided. So the wavelength of light having 2.55 eV will be absorbed.

$\mathrm{\lambda}=\frac{1242}{2.55}=487.05\mathrm{nm}\phantom{\rule{0ex}{0ex}}=487\mathrm{nm}$

The wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength â€‹487 nm will have less intensity in the transmitted beam.

#### Page No 385:

#### Question 28:

Radiation coming from transition *n* = 2 to *n* = 1 of hydrogen atoms falls on helium ions in *n* = 1 and *n* = 2 states. What are the possible transitions of helium ions as they absorbs energy from the radiation?

#### Answer:

Energy of radiation (E) from the hydrogen atom is given by

$E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

Hydrogen atoms go through transition, *n* = 1 to *n* = 2.

The energy released is given by

$E=13.6\left(\frac{1}{1}-\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{3}{4}=10.2\mathrm{eV}$

For He,

Atomic no, *Z* = 2

Let us check the energy required for the transition in helium ions from *n = *1 to *n = *2.

∴ *n*_{1} = 1 to* **n*_{2} = 2

Energy $\left({E}_{1}\right)$ of this transition is given by

${E}_{1}={Z}^{2}13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=4\times 13.6\left(1-\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=40.8\mathrm{eV}$*E*_{1} > *E*,

Hence, this transition of helium ions is not possible.

Let us check the energy required for the transition in helium ion from *n* = 1 to *n* = 3.

∴ *n*_{1} = 1 to *n*_{2} = 3

Energy$\left({E}_{2}\right)$ for this transition is given by

${E}_{2}={Z}^{2}\times 13.6\left(\frac{1}{{{n}_{2}}^{2}}-\frac{1}{{{n}_{1}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=4\times 13.6\times \left(\frac{1}{1}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=48.3\mathrm{eV}$

It is clear that *E*_{2} > *E*.

Hence, this transition of helium ions is not possible.

Similarly, transition from *n*_{1} = 1 to *n*_{2} = 4 is also not possible.

Let us check the energy required for the transition in helium ion from *n* = 2 to *n* = 3.

∴ *n*_{1} = 2 to *n*_{2} = 3

Energy $\left({E}_{3}\right)$ for this transition is given by

${E}_{3}=13.6\times 4\left(\frac{1}{4}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=\frac{20\times 13.6}{36}=7.56\mathrm{eV}$

Let us check the energy required for the transition in helium ion from *n* = 2 to *n* = 3.

∴ *n _{â€‹â€‹}*

_{1 }= 2 to

*n*

_{2}= 4

Energy $\left({E}_{4}\right)$ for this transition is given by

${E}_{4}=13.6\times 4\left(\frac{1}{4}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{3}{4}=10.2\mathrm{eV}$

We find that

*E*

_{3}<

*E*

*E*

_{4}=

*E*

Hence, possible transitions

*are from*

*n*= 2 to

*n*= 3 and

*n*= 2 to

*n*= 4.

#### Page No 385:

#### Question 29:

A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?

#### Answer:

Given:

Wavelength of ultraviolet radiation, *λ* = 50 nm = 50$\times $10^{$-9$}m

We know that the work function of an atom is the energy required to remove an electron from the surface of the atom. So, we can find the work function by calculating the energy required to remove the electron from *n*_{1} = 1 to *n*_{2} = ∞.

Work function, ${W}_{0}=13.6\left(\frac{1}{1}-\frac{1}{\infty}\right)\phantom{\rule{0ex}{0ex}}=13.6\mathrm{eV}$

Using Einstein's photoelectric equation, we get

$E={W}_{0}+\mathrm{KE}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{hc}{\lambda}-13.6=\mathrm{KE}\left(\because E=\frac{hc}{\lambda}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1242}{50}-13.6=\mathrm{KE}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{KE}=24.84-13.6\phantom{\rule{0ex}{0ex}}=11.24\mathrm{eV}$

#### Page No 385:

#### Question 30:

A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.

#### Answer:

Given:

Wavelength of light, *λ* = 100 nm = 100 $\times $ 10^{$-9$} m

Energy of the incident light$\left(E\right)$ is given by*$E=\frac{hc}{\mathrm{\lambda}}$*

Here,*h* = Planck's constant*λ* = Wavelength of light

$\therefore E=\frac{1242}{100}\phantom{\rule{0ex}{0ex}}E=12.42\mathrm{eV}$

(a)

Let *E*_{1}â€‹ and* E*_{2 }be the energies of the 1st and the 2nd state, respectively.

Let the transition take place from* **E*_{1} to* **E*_{2}.

Energy absorbed during this transition is calculated as follows:

Here,*n*_{â€‹â€‹â€‹1} = 1

*n*_{â€‹â€‹â€‹2} = 2

Energy absorbed (*E*') is given by

$E\text{'}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=13.6\left(\frac{1}{1}-\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{3}{4}=10.2\mathrm{eV}$

Energy left = 12.42 eV − 10.2 eV = 2.22 eV

â€‹Energy of the photon = $\frac{hc}{\lambda}$

Equating the energy left with that of the photon, we get

$2.22\mathrm{eV}=\frac{hc}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}2.22\mathrm{eV}=\frac{1242}{\mathrm{\lambda}}$

or** ***λ* = 559.45 = 560 nm

Let *E*_{3}_{}be the energy of the 3rd state.

Energy absorbed for the transition from *E*_{1}_{}to *E*_{3} is given by

$E\text{'}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=13.6\left(\frac{1}{1}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{8}{9}=12.1\mathrm{eV}$

Energy absorbed in the transition from *E*_{1}_{}to *E*_{3} = 12.1 eV (Same as solved above)

Energy left = 12.42 − 12.1 = 0.32 eV

$0.32=\frac{hc}{\mathrm{\lambda}}=\frac{1242}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}\mathrm{\lambda}=\frac{1242}{0.32}\phantom{\rule{0ex}{0ex}}=3881.2=3881\mathrm{nm}$

Let *E*_{4} be the energy of the 4th state.

Energy absorbed in the transition from *E*_{3}_{}to *E*_{4} is given by

$E\text{'}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=13.6\left(\frac{1}{9}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{7}{144}=0.65\mathrm{eV}$

Energy absorbed for the transition from *n* = 3 to *n* = 4 is 0.65 eV

Energy left = 12.42 − 0.65 = 11.77 eV

Equating this energy with the energy of the photon, we get

$11.77=\frac{hc}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{\lambda}=\frac{1242}{11.77}=105.52$

The wavelengths observed in the transmitted beam are 105 nm, 560 nm and 3881 nm.

(b)

If the energy absorbed by the 'H' atom is radiated perpendicularly, then the wavelengths of the radiations detected are calculated in the following way:

$E=10.2\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow 10.2=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{or}\lambda =\frac{1242}{10.2}=121.76\mathrm{nm}\approx 121\mathrm{nm}\phantom{\rule{0ex}{0ex}}E=12.1\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow 12.1=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{or}\lambda =\frac{1242}{12.1}=102.64\mathrm{nm}\approx 103\mathrm{nm}\phantom{\rule{0ex}{0ex}}E=0.65\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.65=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{or}\lambda =\frac{1242}{0.65}=1910.76\mathrm{nm}1911\mathrm{nm}$

Thus, the wavelengths of the radiations detected are 103 nm, 121 nm and 1911 nm.

#### Page No 385:

#### Question 31:

A beam of monochromatic light of wavelength λ ejects photoelectrons from a cesium surface (Φ = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of λ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.

#### Answer:

Given:

Work function of cesium surface, *Ï• *= 1.9 eV

(a) Energy required to ionise a hydrogen atom in its ground state, *E* = 13.6 eV

From the Einstein's photoelectric equation,

$\frac{\mathrm{hc}}{\lambda}=E+\varphi $

Here,* *

h = Planck's constant

c = Speed of light

$\lambda $ = Wavelength of light

$\therefore \frac{\mathrm{hc}}{\mathrm{\lambda}}-1.9=13.6\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1240}{\mathrm{\lambda}}=15.5\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=\frac{1240}{15.5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=80\mathrm{nm}$

(b) When the electron is excited from the states *n*_{1} = 1 to *n*_{2} = 2, energy absorbed $\left({E}_{1}\right)$ is given by

${E}_{1}=13.6\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{E}_{1}=13.6\left(1-\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}{E}_{1}=\frac{13.66\times 3}{4}$

For Einstein's photoelectric equation,

$\therefore \frac{\mathrm{hc}}{\lambda}-1.9=\frac{13.6\times 3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{hc}}{\lambda}=\frac{13.6\times 3}{4}+1.9\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1240}{\lambda}=10.2+1.9=12.1\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{1240}{12.1}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =102.47=102\mathrm{nm}$

(c) Excited atom will emit visible light if an electron jumps from the second orbit_{â€‹ }to third orbit, i.e. from *n*_{1} = 2â€‹ to *n*_{2} = 3. This is because Balmer series lies in the visible region.

Energy (*E*_{2}) of this transition is given by

${E}_{2}=13.6\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{E}_{2}=13.6\left(\frac{1}{4}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}{E}_{2}=\frac{13.66\times 5}{36}$

For Einstein's photoelectric equation,

$\frac{\mathrm{hc}}{\lambda}-1.9=\frac{13.6\times 5}{36}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{hc}}{\lambda}=\frac{13.6\times 5}{36}+1.9\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1240}{\lambda}=1.88+1.9=3.78\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{1240}{3.78}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =328.04\mathrm{nm}$

#### Page No 385:

#### Question 32:

Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?

#### Answer:

Given:

Distance travelled by the electron, *d = * 1.0 m

Wavelength of red light,$\lambda $ = 656.3 nm = 656.3 $\times $ 10^{$-9$} m

Since the given wavelength lies in Balmer series, the transition that requires minimum energy is from *n*_{1} = 3 to *n*_{2} = 2.

Energy of this transition will be equal to the energy $\left({E}_{1}\right)$ that will be required for the transition from the ground state to *n* = 3.

${E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

$\Rightarrow {E}_{1}=13.6\left(1-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=\frac{13.6\times 8}{9}=12.09\mathrm{eV}$

Energy, *E* (e*V*) = 12.09 eV

∴ *V = *12.09 V

Electric field, *E* = $\frac{V}{d}$ = $\frac{12.09}{1}$ = 12.09 V/m

∴ Minimum value of the electric field = 12.09 V/m = 12.1 V/m

#### Page No 385:

#### Question 33:

A neutron having kinetic energy 12.5 eV collides with a hydrogen atom at rest. Nelgect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.

#### Answer:

Given:

Initial kinetic energy of the neutron, *K* = 12.5 eV

The velocities of the two bodies of equal masses undergoing elastic collision in one dimension gets interchanged after the collision.

Since the hydrogen atom is at rest, after collision, the velocity of the neutron will be zero.

Hence, it has zero energy.

#### Page No 385:

#### Question 34:

A hydrogen atom moving at speed υ collides with another hydrogen atom kept at rest. Find the minimum value of υ for which one of the atoms may get ionized.

The mass of a hydrogen atom = 1.67 × 10^{−27} kg.

#### Answer:

Given:

Mass of the hydrogen atom, *M *= 1.67 × 10^{−27} kg

Let* v* be the velocity with which hydrogen atom is moving before collision.

Let *v*_{1} and* **v*_{2} be the velocities of hydrogen atoms after the collision.

Energy used for the ionisation of one atom of hydrogen, Δ*E** = *13.6 eV = 13.6×(1.6×10^{−19})J

Applying the conservation of momentum, we get*mv* = *mv*_{1} + *m*υ_{2} ...(1)

Applying the conservation of mechanical energy, we get

$\frac{1}{2}m{\upsilon}^{2}=\frac{1}{2}m{\upsilon}_{1}^{2}+\frac{1}{2}m{\upsilon}_{2}^{2}+\u2206\mathrm{E}...\left(2\right)$

Using equation (1), we get

*v*^{2} = (*v*_{1} + *v*_{2})^{2}*v ^{2}* $={\upsilon}_{1}^{2}+{\upsilon}_{2}^{2}+2{\upsilon}_{1}{\upsilon}_{2}$ ...(3)

In equation(2), on multiplying both the sides by 2 and dividing both the sides by

*m*.

${\mathrm{\upsilon}}^{2}={\mathrm{\upsilon}}_{1}^{2}+{\mathrm{\upsilon}}_{1}^{2}+2\u2206E/m....\left(4\right)$

On comparing (4) and (3), we get

$\left[\because 2{\upsilon}_{1}{\upsilon}_{2}=\frac{2\u2206E}{m}\right]$

(υ

_{1}− υ

_{2})

^{2}= (υ

_{1}+ υ

_{2})

^{2}− 4υ

_{1}υ

_{2}

_{â€‹${\left({v}_{1}-{v}_{2}\right)}^{2}={v}^{2}-\frac{4\u2206E}{m}$}

For minimum value of υ,

*v*

_{1}

_{ }=

*v*

_{2}

_{ }=0

Also, ${v}^{2}-\frac{4\u2206E}{m}=0$

$\therefore {v}^{2}=\frac{4\u2206E}{m}\phantom{\rule{0ex}{0ex}}=\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}}\phantom{\rule{0ex}{0ex}}={10}^{4}\sqrt{\frac{4\times 13.6\times 1.6}{1.67}}\phantom{\rule{0ex}{0ex}}=7.2\times {10}^{4}\mathrm{m}/\mathrm{s}$

#### Page No 385:

#### Question 35:

A neutron moving with a speed υ strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67 × 10^{−27} kg.

#### Answer:

Given:

Mass of neutron, *m* = 1.67 × 10^{−27} kg

Since neutron is moving with velocity $\left(v\right)$, its energy $\left(E\right)$ is given by

$E=\frac{1}{2}m{\upsilon}^{2}$

Let the energy absorbed be âˆ†*E**.*

The condition for inelastic collision is given below:

$\frac{1}{2}m{\upsilon}^{2}>2\u2206E$

$\Rightarrow \u2206E<\frac{1}{4}m{\upsilon}^{2}$

Since 10.2 eV, energy is required for the first excited state.

∴ âˆ†*E* < 10.2 eV

$\therefore 10.2\mathrm{eV}\frac{1}{4}m{v}^{2}$

Thus, minimum speed of the neutron is given by

$\Rightarrow {\upsilon}_{\mathrm{min}}=\sqrt{\frac{4\times 10.2}{m}}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow {\upsilon}_{\mathrm{min}}=\sqrt{\frac{10.2\times 1.6\times {10}^{19}\times 4}{1.67\times {10}^{-27}}}\phantom{\rule{0ex}{0ex}}=6\times {10}^{4}\mathrm{m}/\mathrm{sec}$

#### Page No 385:

#### Question 36:

When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. (a) Calculate the momentum carries by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom = 1.67 × 10^{−27} kg. (c) Find the kinetic energy of recoil of the atom.

#### Answer:

Given:

Wavelength of light emitted by hydrogen, *λ* = 656.3 nm

Mass of hydrogen atom, *m* = 1.67 × 10^{−27} kg

(a) Momentum $\left(p\right)$ is given by

$P=\frac{h}{\lambda}$

Here,*h* = Planck's constant*λ* = Wavelength of light

$\therefore p=\frac{6.63\times {10}^{-34}}{656.3\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}p=0.01\times {10}^{-25}\phantom{\rule{0ex}{0ex}}p=1\times {10}^{-27}\mathrm{kgm}/\mathrm{s}$

(b) Momentum, *p* = *mv*

Here,*m* = Mass of hydrogen atom*v = *Speed of atom

$\therefore $ 1 × 10^{−27} = (1.67 × 10^{−27})× υ

$\Rightarrow \upsilon =\frac{1}{1.67}\phantom{\rule{0ex}{0ex}}=0.598=0.6\mathrm{m}/\mathrm{s}$

(c) Kinetic energy $\left(K\right)$ of the recoil of the atom is given by

$K=\frac{1}{2}m{v}^{2}$

Here,*m* = Mass of the atom*v* = Velocity of the atom

$\therefore K=\frac{1}{2}\times \left(1.67\times {10}^{-27}\right)\times {\left(0.6\right)}^{2}\mathrm{J}$

$K=\frac{0.3006\times {10}^{-27}}{1.6\times {10}^{-19}}\mathrm{eV}\phantom{\rule{0ex}{0ex}}K=1.9\times {10}^{-9}\mathrm{eV}$

#### Page No 385:

#### Question 37:

When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from *n* = 3 to *n* = 2. Calculate the fractional change in the wavelength of light emitted, due to the recoil.

#### Answer:

Difference in energy in the transition from *n* = 3 to *n* = 2 is 1.89 eV ( = *E*).

If all this energy is used up in emitting a photon (i.e. recoil energy is zero).

Then,

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{hc}{E}...\left(\mathrm{i}\right)$

If difference of energy is used up in emitting a photon and recoil of atom, then let *E _{R}* be the recoil energy of atom.

$E=\frac{hc}{\lambda \text{'}}+{E}_{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda \text{'}=\frac{hc}{E-{E}_{R}}...\left(\mathrm{ii}\right)$

Fractional change in the wavelength is given as,

$\frac{\u2206\lambda}{\lambda}=\frac{\lambda \text{'}-\lambda}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206\lambda}{\lambda}=\frac{1}{\lambda}\left(\frac{hc}{E-{E}_{R}}-\frac{hc}{E}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206\lambda}{\lambda}=\frac{E}{hc}\left(\frac{hc{E}_{R}}{E(E-{E}_{R})}\right)\left(\because \lambda =\frac{hc}{E}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206\lambda}{\lambda}=\left(\frac{{E}_{R}}{(E-{E}_{R}}\right)$

#### Page No 385:

#### Question 38:

The light emitted in the transition *n* = 3 to *n* = 2 in hydrogen is called H_{α} light. Find the maximum work function a metal can have so that H_{α} light can emit photoelectrons from it.

#### Answer:

The H_{α}_{}light can emit the photoelectrons if its energy is greater than or equal to the work function of the metal.

Energy possessed by H_{α} light (*E*) is given by

$E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$ eV

Here, *n*_{1} = 2, *n*_{2} = 3â€‹

$\therefore E=13.6\times \left(\frac{1}{4}-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=\frac{13.6\times 5}{36}=1.89\mathrm{eV}\phantom{\rule{0ex}{0ex}}=1.90\mathrm{eV}$

H_{α} light will be able to emit electron from the metal surface for the maximum work function of metal to be 1.90 eV.

#### Page No 385:

#### Question 39:

Light from Balmer series of hydrogen is able to eject photoelectrons from a metal. What can be the maximum work function of the metal?

#### Answer:

Let the maximum work function of the metal be *W*.

The energy liberated in the Balmer Series (*E*) is given by

$E=13.6\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}$

For maximum work function, maximum energy of Balmer's series is taken.

Now, *n*_{1} = 2, *n*_{1} = ∞â€‹

$\therefore E=13.6\left(\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=13.6\times \frac{1}{4}=3.4\mathrm{eV}$

Here,*W = E*

Thus, maximum work function of metal is 3.4 eV.

#### Page No 385:

#### Question 40:

Radiation from hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. Work function of cesium is 1.9 eV.

#### Answer:

Given:

Work function of cesium, $\varphi $ = 1.9 eVâ€‹

Energy of photons coming from the discharge tube,* E* = 13.6 eV

Let maximum kinetic energy of photoelectrons emitted be *K.*

From the Einstein's photoelectric equation, we know that the maximum kinetic energy of photoelectrons emitted is given by

K = *E* − $\varphi $

= 13.6 eV − 1.9 ev

= 11.7 eVâ€‹

#### Page No 385:

#### Question 41:

A filter transmits only the radiation of wavelength greater than 440 nm. Radiation from a hydrogen-discharge tube goes through such a filter and is incident on a metal of work function 2.0 eV. Find the stopping potential which can stop the photoelectrons.

#### Answer:

Wavelength of radiation coming from filter, *λ* = 440 nm

Work function of metal, *Ï• *= 2 eV

Charge of the electronâ€‹, *e = *1.6 $\times $ 10^{$-9$ }C

Let *V*_{0} be the stopping potential.

From Einstein's photoelectric equation,

$\frac{hc}{\mathrm{\lambda}}-\varphi =e{V}_{0}\phantom{\rule{0ex}{0ex}}$

Here,*h* =Planck constant*c *= Speed of light*$\lambda $* = Wavelength of radiation

$\frac{4.14\times {10}^{-15}\times 3\times {10}^{8}}{440\times {10}^{-9}}-2\mathrm{eV}={\mathrm{eV}}_{0}\phantom{\rule{0ex}{0ex}}\Rightarrow e{V}_{0}=\left(\frac{1242}{440}-2\right)\mathrm{eV}=0.823\mathrm{eV}.\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}=0.823\mathrm{Volts}$

#### Page No 385:

#### Question 42:

The earth revolves round the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principal quantum number *n* for the present radius? Mass of the earth = 6.0 × 10^{−24} kg. Mass of the sun = 2.0 × 10^{30} kg, earth-sun distance = 1.5 × 10^{11} m.

#### Answer:

Given:

Mass of the earth, *m*_{e} = 6.0 × 10^{24} kg

Mass of the sun, *m*_{s} = 2.0 × 10^{30} kg

Distance between the earth and the sun, *d* = 1.5 × 11^{11} m

According to the Bohr's quantization rule,

Angular momentum, *L* =* *$\frac{nh}{2\mathrm{\pi}}$

$\Rightarrow mvr=\frac{nh}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}$ ....(1)

Here,*n = *Quantum number*h = *Planck's constant*m = *Mass of electron*r* = Radius of the circular orbit*v = *Velocity of the electron

Squaring both the sides, we get

${m}_{\mathrm{e}}^{2}{v}^{2}{r}^{2}=\frac{{n}^{2}{h}^{2}}{4{\mathrm{\pi}}^{2}}....\left(2\right)\phantom{\rule{0ex}{0ex}}$

Gravitational force of attraction between the earth and the sun acts as the centripetal force.

$F=\frac{G{m}_{\mathrm{e}}{m}_{\mathrm{s}}}{{r}^{2}}=\frac{{m}_{e}{v}^{2}}{r}$

$\Rightarrow {v}^{2}=\frac{G{m}_{s}}{r}$ ....(3)

Dividing (2) by (3), we get

${m}_{\mathrm{e}}^{2}r=\frac{{n}^{2}{h}^{2}}{4{\mathrm{\pi}}^{2}G{m}_{\mathrm{s}}}$

(a) For *n* = 1,

$r=\sqrt{\frac{{h}^{2}}{4{\mathrm{\pi}}^{2}G{m}_{\mathrm{s}}{m}_{\mathrm{e}}^{2}}}\phantom{\rule{0ex}{0ex}}r=\sqrt{\frac{{\left(6.63\times {10}^{-34}\right)}^{2}}{4\times {\left(3.14\right)}^{2}\times \left(6.67\times {10}^{-11}\right)\times {\left(6\times {10}^{24}\right)}^{2}\times \left(2\times {10}^{30}\right)}}\phantom{\rule{0ex}{0ex}}r=2.29\times {10}^{-138}\mathrm{m}\phantom{\rule{0ex}{0ex}}r=2.3\times {10}^{-138}\mathrm{m}$

(b)

From (2), the value of the principal quantum number (*n*) is given by

${n}^{2}=\frac{{m}_{\mathrm{e}}^{2}\times r\times 4\times \mathrm{\pi}\times \mathrm{G}\times {m}_{\mathrm{s}}}{{h}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\sqrt{\frac{{m}_{\mathrm{e}}^{2}\times r\times 4\times \mathrm{\pi}\times \mathrm{G}\times {m}_{\mathrm{s}}}{{h}^{2}}}$

$n=\sqrt{\frac{{\left(6\times {10}^{24}\right)}^{2}\times \left(1.5\times {10}^{11}\right)\times 4\times {\left(3.14\right)}^{2}\times \left(6.67\times {10}^{-11}\right)\times \left(2\times {10}^{30}\right)}{{\left(6.6\times {10}^{-34}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}n=2.5\times {10}^{74}$

#### Page No 386:

#### Question 43:

Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr's quantization rule for angular momentum to be valid in this case, derive an expression for the energy of the neutron-electron system.

#### Answer:

According to Bohr's quantization rule,

Angular momentum of electron, *L* = $\frac{nh}{2\mathrm{\pi}}$

$\Rightarrow m{v}_{\mathrm{e}}r=\frac{nh}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{e}}=\frac{nh}{2\mathrm{\pi}r{m}_{\mathrm{e}}}$ ...(1)

Here,*n = *Quantum number*h = *Planck's constant*m = *Mass of the electron

*r* = Radius of the circular orbit*v*_{e}* = *Velocity of the electron

Let *m*_{n} be the mass of neutron.

On equating the gravitational force between neutron and electron with the centripetal acceleration,

$\frac{\mathrm{G}{m}_{n}{m}_{e}}{{r}^{2}}=\frac{{m}_{e}{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{G}{m}_{n}}{r}={v}^{2}...\left(2\right)$

Squaring (1) and dividing it by (2), we have

$\frac{{m}_{\mathrm{e}}^{2}{v}^{2}{r}^{2}}{{v}^{2}}=\frac{{n}^{2}{h}^{2}r}{4{\pi}^{2}G{m}_{\mathrm{n}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{\mathrm{e}}^{2}{r}^{2}=\frac{{n}^{2}{h}^{2}r}{4{\mathrm{\pi}}^{2}\mathrm{G}{m}_{\mathrm{n}}}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{{n}^{2}{h}^{2}}{4{\mathrm{\pi}}^{2}\mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}^{2}}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{e}}=\frac{nh}{2\pi r{m}_{\mathrm{e}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{e}}=\frac{nh}{2\pi {m}_{\mathrm{e}}\times \left(\frac{{n}^{2}{h}^{2}}{4{\mathrm{\pi}}^{2}\mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}^{2}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{e}}=\frac{2\pi \mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}}{nh}$

$\mathrm{Kinetic}\mathrm{e}\mathrm{nergy}\mathrm{of}\mathrm{the}\mathrm{electron},K=\frac{1}{2}{m}_{\mathrm{e}}{{v}_{\mathrm{e}}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{m}_{\mathrm{e}}{\left(\frac{2\mathrm{\pi G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}}{nh}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4{\mathrm{\pi}}^{2}{\mathrm{G}}^{2}{m}_{\mathrm{n}}^{2}{m}_{\mathrm{e}}^{3}}{2{n}^{2}{h}^{2}}\phantom{\rule{0ex}{0ex}}$

Potential energy of the neutron, $P=\frac{-\mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}}{r}$

Substituting the value of *r* in the above expression,

$P=\frac{-\mathrm{G}{m}_{\mathrm{e}}{m}_{\mathrm{n}}4{\mathrm{\pi}}^{2}\mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}^{2}}{{n}^{2}{h}^{2}}\phantom{\rule{0ex}{0ex}}P=\frac{-4{\mathrm{\pi}}^{2}{\mathrm{G}}^{2}{m}_{\mathrm{n}}^{2}{m}_{\mathrm{e}}^{3}}{{n}^{2}{h}^{2}}$

Total energy = *K + P*$=-\frac{2{\mathrm{\pi}}^{2}{\mathrm{G}}^{2}{m}_{n}^{2}{m}_{e}^{2}}{2{n}^{2}{h}^{2}}=-\frac{{\mathrm{\pi}}^{2}{\mathrm{G}}^{2}{m}_{n}^{2}{m}_{e}^{2}}{{n}^{2}{h}^{2}}$

#### Page No 386:

#### Question 44:

A uniform magnetic field B exist in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the minimum possible speed of the electron.

#### Answer:

According to Bohr's quantization rule,

$mvr=\frac{nh}{2\mathrm{\pi}}$

'*r*' is minimum when '*n*' has minimum value, i.e. 1.

$mv=\frac{nh}{2\mathrm{\pi}r}...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow mv=rq\mathrm{B}...\left(2\right)$

From (1) and (2), we get

$rq\mathrm{B}=\frac{nh}{2\mathrm{\pi}r}\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{nh}{2\mathrm{\pi}e\mathrm{B}}\left[\therefore q=e\right]\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{\frac{h}{2\mathrm{\pi eB}}}\left[n=1\right]$

(b) For the radius of *n*^{th} orbit,

$r=\sqrt{\frac{nh}{2\mathrm{\pi}e\mathrm{B}}}$

(c) $mvr=\frac{nh}{2\mathrm{\pi}},r=\frac{mv}{q\mathrm{B}}$

Substituting the value of '*r*' in (1), we get

$mv\times \frac{mv}{q\mathrm{B}}=\frac{nh}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}^{2}{v}^{2}=\frac{he\mathrm{B}}{2\mathrm{\pi}}\left[n=1,q=e\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}=\frac{he\mathrm{B}}{2\mathrm{\pi}{m}^{2}}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{he\mathrm{B}}{2\mathrm{\pi}{m}^{2}}}$

#### Page No 386:

#### Question 45:

Suppose in an imaginary world the angular momentum is quantized to be even integral multiples of *h*/2π. What is the longest possible wavelength emitted by hydrogen atoms in visible range in such a world according to Bohr's model?

#### Answer:

In the imaginary world, the angular momentum is quantized to be an even integral multiple of *h*/2$\mathrm{\pi}$.

Therefore, the quantum numbers that are allowed are *n*_{1} = 2 and *n*_{2} = 4.

We have the longest possible wavelength for minimum energy.

Energy of the light emitted (*E*) is given by

$E=13.6\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}E=13.6\left[\frac{1}{{\left(2\right)}^{2}}-\frac{1}{{\left(4\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}E=13.6\left(\frac{1}{4}-\frac{1}{16}\right)\phantom{\rule{0ex}{0ex}}E=\frac{13.6\times 12}{64}=2.55\mathrm{eV}$

Equating the calculated energy with that of photon, we get

$2.55\mathrm{eV}=\frac{hc}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}\mathrm{\lambda}=\frac{hc}{2.55}=\frac{1242}{2.55}\mathrm{nm}\phantom{\rule{0ex}{0ex}}=487.05\mathrm{nm}=487\mathrm{nm}$

#### Page No 386:

#### Question 46:

Consider an excited hydrogen atom in state *n* moving with a velocity υ(ν<<*c*). It emits a photon in the direction of its motion and changes its state to a lower state *m*. Apply momentum and energy conservation principles to calculate the frequency ν of the emitted radiation. Compare this with the frequency ν_{0} emitted if the atom were at rest.

#### Answer:

Let the frequency emitted by the atom at rest be *ν*_{0}.

Let the velocity of hydrogen atom in state '*n*' be *u*.

But *u* << *c*

Here, the velocity of the emitted photon must be *u*.

According to the Doppler's effect,

The frequency of the emitted radiation, *ν* is given by

$\mathrm{Frequency}\mathrm{of}\mathrm{the}\mathrm{emitted}\mathrm{radiation},\nu ={\nu}_{0}\left(\frac{1+{\displaystyle \frac{u}{c}}}{1-{\displaystyle \frac{u}{c}}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Since}uc,\phantom{\rule{0ex}{0ex}}\nu ={\nu}_{0}\left(\frac{1+{\displaystyle \frac{u}{c}}}{1}\right)\phantom{\rule{0ex}{0ex}}\nu ={\nu}_{0}\left(1+\frac{u}{c}\right)$

Ratio of frequencies of the emitted radiation, $\frac{\nu}{{\nu}_{0}}=\left(1+\frac{u}{c}\right)$

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