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#### Page No 382:

Balmer series contains wavelengths ranging from 364 nm (for n2 = 3) to 655 nm (n2 = $\infty$).
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 = 1.097×107 m$-$1

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

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

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

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

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

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 n2 = 2) to 121 nm (n2 =$\infty$).
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

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

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

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

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

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

#### Page No 382:

The energy of hydrogen ion is given by

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:

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

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:

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:

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 n2 =$\infty$) to 655 nm (for n2 =3).

#### Page No 383:

Energy of nth state of hydrogen is given by

Energy of first excited state (n = 2) of hydrogen, E1 = = -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,
E1, = -3.4 eV-10 eV = -13.4 eV
We still write En = E1/n2, or rn = a0 n2 because these formulas are independent of the refrence point enegy.

#### Page No 383:

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

The frequency of the radiation emitted for transition from n1 to n2  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,
n1 = 1
n2 = $\infty$

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

For the first line of Lyman series,
n1 = 1
n2 = 2

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

For series limit of Balmer series,
n1 = 2
n2 = $\infty$

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

f1 $-$ f3 = f2

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:

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:

When a photon of energy (E2E1 = hυ) is incident on an atom in the ground state, the atom in the ground state E1 may absorb the photon and jump to a higher energy state (E2). 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:

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

For microwave region

This implies that stimulated transition dominate in this region.
For visible region

So here spontaneous transition dominate.

#### Page No 383:

(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:

(d) either two atoms or three atoms

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

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:

(c) 10−42 N-m

The angular momentum of the electron for the nth 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.

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

#### Page No 383:

(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:

(d) Doubly ionized lithium

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

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

Here, a0 = 0.53 pm

For lithium,
Z = 3
Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

#### Page No 383:

(d) Doubly ionized lithium

The wavelength corresponding the transition from n2 to n1 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:

(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:

(b) increases

The electric potential energy of hydrogen atom with electron at the nth state is given by
$V=-\frac{2×13.6}{{n}^{2}}$
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:

(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 Z2 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$-$ 13.6×22$-$ 54.4 eV

For Li++,
Z = 3
∴ Total energy, E$-$ 13.6×32 = $-$ 122.4 eV

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

#### Page No 383:

(d) Li++

The radius of the nth orbit in one electron system is given by
${r}_{\mathrm{n}}=\frac{{n}^{2}{a}_{0}}{Z}$

Here, a0 = 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 =

For Li++,
Z = 3

∴ Radius of Li++ atom =

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

#### Page No 383:

(a) 1.05 × 10−34 J s

Let after absorption of energy, the hydrogen atom goes to the nth excited state.

Therefore, the energy absorbed can be written as
$10.2=13.6×\left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{10.2}{13.6}=1-\frac{1}{{n}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{n}^{2}}=\frac{13.6-10.2}{13.6}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{n}^{2}}=\frac{3.4}{13.6}\phantom{\rule{0ex}{0ex}}⇒{n}^{2}=4\phantom{\rule{0ex}{0ex}}⇒n=2\phantom{\rule{0ex}{0ex}}$

The orbital angular momentum of the electron in the nth state is given by
${L}_{\mathrm{n}}=\frac{nh}{2\pi }$
Change in the angular momentum, $∆L=\frac{2h}{2\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }$

#### Page No 383:

(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:

(d) move with same speed

All the photons emitted in the laser move with the speed equal to the speed of light (c = 3×108 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:

(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:

(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:

(a) vn
(b) Er
Relations for energy, radius of the orbit and its velocity are given by

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, Er.

#### Page No 384:

(a) will pass through the origin
(b) will be a straight line with slope 4

The radius of the nth orbit of a hydrogen atom is given by
${r}_{\mathrm{n}}={n}^{2}{a}_{0}$
Area of the nth orbit is given by

From the above expression, the graph of ln (An/A1) against ln(n) will be a straight line passing through the origin and having slope 4.

#### Page No 384:

(b) uA > uB
The ionisation energy of a hydrogen like ion of atomic number Z is given by

Thus, the atomic number of ion A is greater than that of B (ZA > ZB).
The radius of the orbit is inversely proportional to the atomic number of the ion.
rA > rB
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, uA > uB 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 Z2, the energy of A will be less than that of B, i.e.  EA < EB.
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation LA > LB is invalid.

#### Page No 384:

(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:

The dimensions of ε0 can be derived from the formula given below:

Clearly, a0 has the dimensions of length.

#### Page No 384:

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$×$10m$-1$
n1 = Quantum number of final state
n2 = Quantum number of initial state
(a)
For transition from n = 3 to n = 2:

Here,
n1 = 2
n2 = 3

(b)
For transition from n = 5 to n = 4:

Here,
n1 = 4
n2 = 5

(c)
For transition from n = 10 to n = 9:

Here,
n1 = 9
n2 = 10

#### Page No 384:

Given:
For the smallest wavelength, energy should be maximum.
Thus, for maximum energy, transition should be from infinity to the ground state.

n1 = 1
n2$\infty$

(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 = 1.097×107 m$-$1

On substituting the respective values,

(b)
For He+,
Atomic number, Z = 2
Wavelength of the radiation emitted by He($\lambda$) is given by
$\frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

(c) For Li++,
Atomic number, Z = 3
Wavelength of the radiation emitted by  Li++ ($\lambda$) is given by
$\frac{\mathit{1}}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

#### Page No 384:

Expression of Rydberg constant (R) is given by

Mass of electron, me = 9.31$×$1021 kg
Charge, e = 1.6 × 10−19 C
Planck's constant, h = 6.63 × 10−34 J-s,
Speed of light, c = 3 × 108 m/s,
Permittivity of vacuum, ∈0 = 8.85 × 10−12  C2N$-1$m
On substituting the values in the expression, we get

#### Page No 384:

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

For state n = 2,

Thus, binding energy of hydrogen at n = 2 is

#### Page No 384:

For He+ ion,
Atomic number, Z = 2
For hydrogen like ions, radius $\left(r\right)$ of the nth state is given by
Å
Here,
Z = Atomic number of ions
n = Quantum number of the state

Energy $\left(E\right)$ of the nth state is given by
En =
(a)
For n = 1,

Energy, En$\frac{-13.6×4}{1}$
= $-$ 54.4 eV

(b)
For n = 4,
Energy,
(c)
For n = 10,
Radius, $r=\frac{0.53×100}{2}$
= 26.5 Å
Energy,

#### Page No 384:

Wavelength of ultraviolet radiation, $\lambda$ = 102.5 nm = 102,5 $×$10$-9$ m
Rydberg's constant, R = 1.097 107 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= 1) from any excited state (n​2).
Wavelength of light $\left(\lambda \right)$ is given by

The transition will be from 1 to 3.

#### Page No 384:

(a) PE of hydrogen like atom in the nth 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+

⇒10.2 × Z2
= 10.2 × 4
= 40.8 V

(b) Ionization Potential Li++ = 13.6 V × Z2
= 13.6 × 9
= 122.4 V

#### Page No 384:

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$1) be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state.​
Here,
n1 = 2
n2 = 4
Now, the wavelength ($\lambda$1)​ will be

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

Similarly, wavelength ($\lambda$3) for the transition from (n = 3) to (= 2) is given by
When the transition is n1 = 2 to n2 = 3:

#### Page No 384:

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

As the transition takes place from n = 1 to n = 2, the excitation energy (E1) will be

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

The ion may be helium.

#### Page No 384:

Charge on the electron, q1 = 1.6$×$10 $-19$
Charge on the nucleus, q2 = 1.6$×$10 $-19$

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, q1 = q2 = q = 1.6$×$10 $-19$
Smallest distance between the nucleus and the first orbit, r = 0.53$×$10$-$10 m
$K=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}$ = 9$×$109 Nm2C$-$2
Substituting the respective values in (1), we get

#### Page No 384:

(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,
${{n}_{1}}^{2}=\frac{13.6}{10.2}\phantom{\rule{0ex}{0ex}}{n}_{1}=1.15\phantom{\rule{0ex}{0ex}}⇒{n}_{1}=2\phantom{\rule{0ex}{0ex}}$
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
n1 and n2 are quantum numbers.

#### Page No 384:

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 1.097×107 m-1.

#### Page No 384:

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

For n = 6,

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)

(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 nth state can expressed as

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

#### Page No 384:

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).

Therefore, the excitation energy is given by

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:

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,
= 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

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.

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

#### Page No 384:

(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:

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

#### Page No 384:

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 + ∆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:

Given:
Possible transitions:
From n1 = 1 to n2 = 3
n1 = 2 to n2 = 4

(a) Here, n1 = 1 and n2 = 3
Energy, $E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

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

(b) Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.

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

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 λ1 = 487 nm

#### Page No 384:

Let v0 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
$⇒$ $\lambda =\frac{c}{f}$

#### Page No 385:

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
$⇒$1.5 × 8.62 × 10−5 × T = 13.6

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

#### Page No 385:

Given:
Wavelength of red light, $\lambda$ = 653.1 nm = 653.1$×$10$-9$ m

Kinetic energy of H2 molecules $\left(K\right)$ is given by
...(1)
Here, k = 8.62 × 10−5 eV/K
T = Temperature of H2 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, n1 = 1
Also, n2 = 3

Kinetic energy of H2 molecules = Energy released when hydrogen atom goes from ground state to n = 3 state

#### Page No 385:

Frequency of electron (f) is given by

Time period is given by

Here,
h = Planck's constant
m = Mass of the electron
e = Charge on the electron
${\epsilon }_{0}$= Permittivity of free space

Average life time of hydrogen, t = 10$-$8 s
Number of revolutions is given by
$N=\frac{t}{T}\phantom{\rule{0ex}{0ex}}⇒N=\frac{{10}^{-8}}{12247.735×{10}^{-19}}$
N = 8.2 $×$ 105 revolution

#### Page No 385:

Mass of the electron, m = 9.1×10$-$31kg
Radius of the ground state, r = 0.53×10$-$10m
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 = niA = qfA

Here,
h = Planck's constant
e =  Charge on the electron
${\epsilon }_{0}$ = Permittivity of free space
n = Principal quantum number

#### Page No 385:

Mass of the electron, m = 9.1×10$-$31kg
Radius of the ground state, r = 0.53×10$-$10m
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 = niA = qfA

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

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:

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
E1 = $\frac{hc}{{\lambda }_{1}}$
Here,
c = Speed of light
h = Planck's constant

= 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
E2 = $\frac{hc}{{\lambda }_{2}}$

The given range of wavelengths lies in the visible range.
n1 = 2, n2 = 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.

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

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

#### Page No 385:

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

For He,
Atomic no, Z = 2
Let us check the energy required for the transition in helium ions from n = 1 to n = 2.

n1 = 1 to n2 = 2

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

E1 > 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.
n1 = 1 to n2 = 3

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

It is clear that E2 > E.
Hence, this transition of helium ions is not possible.

Similarly, transition from n1 = 1 to n2 = 4 is also not possible.

Let us check the energy required for the transition in helium ion from n = 2 to n = 3.
n1 = 2 to n2 = 3
Energy $\left({E}_{3}\right)$ for this transition is given by

Let us check the energy required for the transition in helium ion from n = 2 to n = 3.
n​​= 2 to n2 = 4
Energy $\left({E}_{4}\right)$ for this transition is given by

We find that
E3 < E
E4 = E
Hence, possible transitions are from n = 2 to n = 3 and n = 2 to n = 4.

#### Page No 385:

Given:
Wavelength of ultraviolet radiation, λ = 50 nm = 50$×$10$-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 n1 = 1 to n2 = ∞.

Work function,
Using Einstein's photoelectric equation, we get

#### Page No 385:

Given:
Wavelength of light, λ = 100 nm = 100 $×$ 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

(a)
Let E1​ and E2 be the energies of the 1st and the 2nd state, respectively.
Let the transition take place from E1 to E2.
Energy absorbed during this transition is calculated as follows:
Here,
n​​​1 = 1
n​​​2 = 2
Energy absorbed (E') is given by

Energy left = 12.42 eV − 10.2 eV = 2.22 eV
​Energy of the photon = $\frac{hc}{\lambda }$
Equating the energy left with that of the photon, we get

or λ = 559.45 = 560 nm
Let E3 be the energy of the 3rd state.
Energy absorbed for the transition from E1 to E3 is given by

Energy absorbed in the transition from E1 to E3 = 12.1 eV  (Same as solved above)

Energy left = 12.42 − 12.1 = 0.32 eV

Let E4 be the energy of the 4th state.
Energy absorbed in the transition from E3 to E4 is given by

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

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:

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

#### Page No 385:

Given:

Work function of cesium surface, ϕ = 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

(b) When the electron is excited from the states n1 = 1 to n2 = 2, energy absorbed $\left({E}_{1}\right)$ is given by
${E}_{1}=13.6\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{E}_{1}=13.6\left(1-\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}{E}_{1}=\frac{13.66×3}{4}$

For Einstein's photoelectric equation,

(c) Excited atom will emit visible light if an electron jumps from the second orbitto third orbit, i.e. from n1 = 2​ to  n2 = 3. This is because Balmer series lies in the visible region.
Energy (E2) of this transition is given by

For Einstein's photoelectric equation,

#### Page No 385:

Given:
Distance travelled by the electron, d = 1.0 m
Wavelength of red light, = 656.3 nm = 656.3 $×$ 10$-9$ m

Since the given wavelength lies in Balmer series, the transition that requires minimum energy is from n1 = 3 to n2 = 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.
${E}_{1}=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$

Energy, E (eV) = 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:

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:

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 v1 and v2 be the velocities of hydrogen atoms after the collision.
Energy used for the ionisation of one atom of hydrogen, ΔE = 13.6 eV = 13.6×(1.6×10−19)J
Applying the conservation of momentum, we get
mv = mv1 + mυ2   ...(1)

Applying the conservation of mechanical energy, we get

Using equation (1), we get
v2 = (v1 + v2)2
v2 $={\upsilon }_{1}^{2}+{\upsilon }_{2}^{2}+2{\upsilon }_{1}{\upsilon }_{2}$           ...(3)

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

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

1 − υ2)2 = (υ1 + υ2)2 − 4υ1υ2
${\left({v}_{1}-{v}_{2}\right)}^{2}={v}^{2}-\frac{4∆E}{m}$

For minimum value of υ,
v1 = v2 =0
Also, ${v}^{2}-\frac{4∆E}{m}=0$

#### Page No 385:

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 ∆E.
The condition for inelastic collision is given below:
$\frac{1}{2}m{\upsilon }^{2}>2∆E$
$⇒∆E<\frac{1}{4}m{\upsilon }^{2}$

Since 10.2 eV, energy is required for the first excited state.
∴ ∆E < 10.2 eV

Thus, minimum speed of the neutron is given by

#### Page No 385:

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

(b) Momentum, p = mv
Here,
m = Mass of hydrogen atom
v = Speed of atom
$\therefore$ 1 × 10−27 = (1.67 × 10−27)× υ

(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

#### Page No 385:

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,

If difference of energy is used up in emitting a photon and recoil of atom, then let ER be the recoil energy of atom.

Fractional change in the wavelength is given as,

#### Page No 385:

The Hα light can emit the photoelectrons if its energy is greater than or equal to the work function of the metal.
Energy possessed by Hα light (E) is given by
$E=13.6\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)$ eV
Here, n1 = 2, n2 = 3​

Hα 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:

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, n1 = 2, n1 = ∞​

Here,
W = E
Thus, maximum work function of metal is 3.4 eV.

#### Page No 385:

Given:
Work function of cesium, $\varphi$ = 1.9 eV​
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 eV​

#### Page No 385:

Wavelength of radiation coming from filter, λ = 440 nm
Work function of metal, ϕ = 2 eV
Charge of the electron​, e = 1.6 $×$ 10$-9$ C
Let V0 be the stopping potential.
From Einstein's photoelectric equation,

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

#### Page No 385:

Given:
Mass of the earth, me = 6.0 × 1024 kg
Mass of the sun, ms = 2.0 × 1030 kg
Distance between the earth and the sun, d = 1.5 × 1111 m

According to the Bohr's quantization rule,

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

Gravitational force of attraction between the earth and the sun acts as the centripetal force.
$F=\frac{G{m}_{\mathrm{e}}{m}_{\mathrm{s}}}{{r}^{2}}=\frac{{m}_{e}{v}^{2}}{r}$
$⇒{v}^{2}=\frac{G{m}_{s}}{r}$     ....(3)
Dividing (2) by (3), we get

(a) For n = 1,

(b)
From (2), the value of the principal quantum number (n) is given by
${n}^{2}=\frac{{m}_{\mathrm{e}}^{2}×r×4×\mathrm{\pi }×\mathrm{G}×{m}_{\mathrm{s}}}{{h}^{2}}\phantom{\rule{0ex}{0ex}}⇒n=\sqrt{\frac{{m}_{\mathrm{e}}^{2}×r×4×\mathrm{\pi }×\mathrm{G}×{m}_{\mathrm{s}}}{{h}^{2}}}$
$n=\sqrt{\frac{{\left(6×{10}^{24}\right)}^{2}×\left(1.5×{10}^{11}\right)×4×{\left(3.14\right)}^{2}×\left(6.67×{10}^{-11}\right)×\left(2×{10}^{30}\right)}{{\left(6.6×{10}^{-34}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}n=2.5×{10}^{74}$

#### Page No 386:

According to Bohr's quantization rule,

Angular momentum of electron,  L = $\frac{nh}{2\mathrm{\pi }}$
$⇒m{v}_{\mathrm{e}}r=\frac{nh}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒{v}_{\mathrm{e}}=\frac{nh}{2\mathrm{\pi }r{m}_{\mathrm{e}}}$          ...(1)
Here,
n = Quantum number
h  = Planck's constant
m = Mass of the electron
r = Radius of the circular orbit
ve = Velocity of the electron

Let mn be the mass of neutron.

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

Squaring (1) and dividing it by (2), we have
$\frac{{m}_{\mathrm{e}}^{2}{v}^{2}{r}^{2}}{{v}^{2}}=\frac{{n}^{2}{h}^{2}r}{4{\pi }^{2}G{m}_{\mathrm{n}}}\phantom{\rule{0ex}{0ex}}⇒{m}_{\mathrm{e}}^{2}{r}^{2}=\frac{{n}^{2}{h}^{2}r}{4{\mathrm{\pi }}^{2}\mathrm{G}{m}_{\mathrm{n}}}\phantom{\rule{0ex}{0ex}}⇒r=\frac{{n}^{2}{h}^{2}}{4{\mathrm{\pi }}^{2}\mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}^{2}}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{e}}=\frac{nh}{2\pi r{m}_{\mathrm{e}}}\phantom{\rule{0ex}{0ex}}⇒{v}_{\mathrm{e}}=\frac{nh}{2\pi {m}_{\mathrm{e}}×\left(\frac{{n}^{2}{h}^{2}}{4{\mathrm{\pi }}^{2}\mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}^{2}}\right)}\phantom{\rule{0ex}{0ex}}⇒{v}_{\mathrm{e}}=\frac{2\pi \mathrm{G}{m}_{\mathrm{n}}{m}_{\mathrm{e}}}{nh}$

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:

According to Bohr's quantization rule,

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

From (1) and (2), we get

(b) For the radius of nth orbit,
$r=\sqrt{\frac{nh}{2\mathrm{\pi }e\mathrm{B}}}$

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

#### Page No 386:

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 n1 = 2 and n2 = 4.

We have the longest possible wavelength for minimum energy.
Energy of the light emitted (E) is given by

Equating the calculated energy with that of photon, we get

#### Page No 386:

Let the frequency emitted by the atom at rest be ν0.

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

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

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