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#### Page No 75:

#### Question 1:

Taking the Bohr radius as* a*_{0} = 53 pm, the radius of Li^{++} ion in its ground state, on the basis of Bohr’s model, will be about

(a) 53 pm

(b) 27 pm

(c) 18 pm

(d) 13 pm

#### Answer:

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

According to Bohr's model, radius of an atom ,

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

The atomic number of lithium, Z = 3,

therefore, the radius of Li^{+ +} ion in its ground state (*n* = 1),

$\Rightarrow {a}_{1}=\left(53\times \frac{{1}^{2}}{3}\right)\mathrm{pm}\approx 18\mathrm{pm}$

Hence, the correct answer is option (c).

#### Page No 75:

#### Question 2:

$\mathrm{B}=-\frac{M{e}^{4}}{8{n}^{2}{e}_{0}^{2}{h}^{2}}$ (

*M*= proton mass)

This last expression is not correct because

(a)

*n*would not be integral.

(b) Bohr-quantisation applies only to electron

(c) the frame in which the electron is at rest is not inertial.

(d) the motion of the proton would not be in circular orbits, even approximately.

#### Answer:

When one decides to work in a frame of reference where the electron is at rest, the given expression is not true as the electron is in non- inertial frame. And in non- inertial frame the same expression is not valid.

Hence, the correct answer is option (c).

#### Page No 76:

#### Question 3:

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because

(a) of the electrons not being subject to a central force.

(b) of the electrons colliding with each other

(c) of screening effects

(d) the force between the nucleus and an electron will no longer be given by Coulomb’s law.

#### Answer:

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because all the electrons present in the atoms are not being subjected to one single central force.

â€‹Hence, the correct answer is option (a).

#### Page No 76:

#### Question 4:

For the ground state, the electron in the H-atom has an angular momentum = â„, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

(a) because Bohr model gives incorrect values of angular momentum.

(b) because only one of these would have a minimum energy.

(c) angular momentum must be in the direction of spin of electron.

(d) because electrons go around only in horizontal orbits.

#### Answer:

Bohr model, does not give the correct values of angular momentum of revolving electrons. According to Bohr's model,

angular momentum of an electron can only be the integral multiple of $\frac{h}{2\pi}$ (*h* = Plank's constant).

Hence, the correct answer is option (a).

#### Page No 76:

#### Question 5:

O_{2} molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

(a) is not important because nuclear forces are short-ranged.

(b) is as important as electrostatic force for binding the two atoms.

(c) cancels the repulsive electrostatic force between the nuclei.

(d) is not important because oxygen nucleus have equal number of neutrons and protons.

#### Answer:

Nuclear force is a short-range force and is operative only over the size of nucleus.

In the oxygen molecules molecules, nuclear force between the nuclei of the two atoms is not important because the distance between the nuclei of the atoms in oxygen molecule is out of the nuclear range.

Hence, the correct answer is option (a).

#### Page No 76:

#### Question 6:

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

(a) 10.20 eV

(b) 20.40 eV

(c) 13.6 eV

(d) 27.2 eV

#### Answer:

The kinetic energy associated with the two H-atoms in the ground state collide inelastically

= 2 × (13.6 eV) = 27.2 eV.

The maximum amount by which their combined kinetic energy is reduced when any one of them goes into first excited state after the inelastic collision.

The total kinetic energy associated with the two H-atoms after the collision

$=\left(\frac{13.6}{{2}^{2}}\right)+\left(13.6\right)=17.0\mathrm{eV}$

Therefore, maximum loss of their combined kinetic energy = 27.2 − 17.0 = 10.20 eV.

Hence, the correct answer is option (a).

#### Page No 76:

#### Question 7:

A set of atoms in an excited state decays.

(a) in general to any of the states with lower energy.

(b) into a lower state only when excited by an external electric field.

(c) all together simultaneously into a lower state.

(d) to emit photons only when they collide.

#### Answer:

A set of atoms in an excited state decays in general to any of the states with lower energy which is more stable than the excited state.

â€‹

Hence, the correct answer is option (a).

#### Page No 77:

#### Question 8:

An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,

(a) the electron would not move in circular orbits.

(b) the energy would be (2)^{4} times that of a H-atom.

(c) the electrons, orbit would go around the protons.

(d) the molecule will soon decay in a proton and a H-atom.

#### Answer:

In an ionised hydrogen molecules, as there are two protons and one electron, therefore, electron's orbit would go around the protons separated by a small distance. This orbit shall not be a circular orbit.

Hence, the correct options are (a) and (c).

#### Page No 77:

#### Question 9:

Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

(a) because of energy conservation.

(b) without simultaneously releasing energy in the from of radiation.

(c) because of momentum conservation.

(d) because of angular momentum conservation.

#### Answer:

The binding energy of H - atom is larger. Large amount of energy is required so that electron can reach near the proton to form H-atom.

When a beam of free electrons is aimed towards free protons, the electrons get scattered on account of energy conservation. An electron and a proton can combine to produce a H-atom only if they simultaneously release energy in the form of radiation.

Hence, the correct options are (a) and (b).

#### Page No 77:

#### Question 10:

The Bohr model for the spectra of a H-atom

(a) will not be applicable to hydrogen in the molecular from.

(b) will not be applicable as it is for a He-atom.

(c) is valid only at room temperature.

(d) predicts continuous as well as discrete spectral lines.

#### Answer:

The Bohr model is valid for only H-atom or H-like atom, i.e., for one electron system only. So, the Bohr model for the spectra of a H-atom will not be applicable to hydrogen in the molecular form (H_{2}). And also, it will not be applicable as it is for a He-atom.

Hence, the correct answer is option (a) and (b).

#### Page No 77:

#### Question 11:

The Balmer series for the H-atom can be observed

(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state.

(b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state.

(c) in any transition in a H-atom.

(d) as a sequence of frequencies with the higher frequencies getting closely packed.

#### Answer:

Balmer series for the H-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states

and the first excited state and as a sequence of frequencies with the higher frequencies getting closely packed.

Hence, the correct answer is option B and D.

#### Page No 77:

#### Question 12:

Let E* _{n}$=\frac{-1}{8{\epsilon}_{0}^{2}}\frac{m{e}^{4}}{{n}^{2}{h}^{2}}$ *be the energy of the

*n*

^{th}level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E

_{2}-E

_{1})/

*h*falls on it,

(a) it will not be absorbed at all

(b) some of atoms will move to the first excited state.

(c) all atoms will be excited to the

*n*= 2 state.

(d) no atoms will make a transition to the

*n*= 3 state.

#### Answer:

When all the H-atoms are in the ground state and radiation of frequency $\left(\frac{{E}_{2}-{E}_{1}}{h}\right)$ falls on it, some of atoms will move to the first excited state and no atoms will make a transition to the n = 3 state.

Hence, the correct answer is option B and D.

#### Page No 78:

#### Question 13:

The simple Bohr modle is not applicable to He^{4} atom because

(a) He^{4} is an inert gas.

(b) He^{4 }has neutrons in the nucleus.

(c) He^{4} has one more electron.

(d) electrons are not subject to central forces.

#### Answer:

The simple Bohr model is not applicable to He^{ 4} atom because He^{4} has one more electron and electrons are not subject to central forces.

Hence, the correct answer is option C and D.

#### Page No 78:

#### Question 14:

The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?

#### Answer:

Since, the difference in mass of a nucleus and its constituents, âˆ†M, is called the mass defect and is given by

$\u2206M=\left[Z{m}_{p}+\left(A-Z\right){m}_{n}\right]-M$

Also, the binding energy is given by,*B* = $\u2206M{c}^{2}$ ≈ 13.6 eV

Thus, the mass of a H-atom is ${m}_{p}+{m}_{e}-\frac{B}{{c}^{2}}$.

Hence, the mass of a H-atom is less than the sum of the masses of a proton and electron.

#### Page No 78:

#### Question 15:

Imagine removing one electron from He^{4} and He^{3}. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

#### Answer:

On removing one electron from He^{4} and He^{3 }, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass. Also after removing one electron from He^{ 4} and He^{ 3} atoms contain one electron and are hydrogen like atoms.

#### Page No 78:

#### Question 16:

When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy?

#### Answer:

The transition of an electron from a higher energy to a lower energy level can appears in the form of electromagnetic radiation because electrons interact only electromagnetically

#### Page No 78:

#### Question 17:

Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)*e* and electron a charge (−3/4)*e *,

where *e* = 1.6 × 10^{–19}C. Give reasons for your answer.

#### Answer:

f proton had a charge $\left(+\frac{4}{3}e\right)$ and electron a charge $\left(-\frac{3}{4}e\right)$ , then the Bohr formula for the H-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

#### Page No 78:

#### Question 18:

Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

#### Answer:

According to Bohr model electrons having different energies belong to different levels having different values of *n*. So, their angular momenta will be different, as

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

#### Page No 78:

#### Question 19:

Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

#### Answer:

The total energy of the electron in the stationary states of the hydrogen atom is given by

${E}_{n}=-\frac{m{e}^{4}}{8{n}^{2}{{\epsilon}_{o}}^{2}{h}^{2}}$

where symbols have their usual meaning and Also, the total energy of the electron in the ground state of the hydrogen atom is −13.6 eV.

For H-atom reduced mass *m*_{e} .

The *m* is the reduced mass.

Whereas for positronium, the reduced mass is,

$m\text{'}\approx \frac{{m}_{e}}{2}$

Hence, the total energy of the electron in the ground state of the positronium atom is:

$E\text{'}=\frac{-13.6}{2}\mathrm{eV}$ = $-6.8$ eV.

#### Page No 78:

#### Question 20:

Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.

#### Answer:

For a He - nucleus with charge 2e and electrons of charge − e, the energy level in ground state is

$-{E}_{n}={Z}^{2}\frac{-13.6\mathrm{eV}}{{n}^{2}}={2}^{2}\frac{-13.6\mathrm{eV}}{{1}^{2}}=-54.4\mathrm{eV}$

Thus, the ground state will have two electrons each of energy E and the total ground state energy would be $-\left(4\times 13.6\right)=-54.4\mathrm{eV}$

#### Page No 79:

#### Question 21:

Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

#### Answer:

The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius $\left({a}_{n}\right)$. Let the velocity of electron is $v$.

Number of revolutions per unit time = $\frac{2\pi {a}_{o}}{v}$

The electric current is given by $I=\frac{q}{t}$, if *q* is charges flows in time *t. *Here *q* = e.

The electric current is given by $I=\frac{2\pi {a}_{o}}{v}e$

#### Page No 79:

#### Question 22:

Show that the first few frequencies of light that is emitted when electrons fall to nth level from levels higher than *n*, arc approximate harmonics (i.e., in the ratio 1 2 3: : K) when* n* >> 1.

#### Answer:

The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from ( ) n p + level to *n*^{th} level can be expressed as a difference of two terms:

${v}_{mn}=cR{Z}^{2}\left[\frac{1}{{\left(n+p\right)}^{2}}-\frac{1}{{n}^{2}}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{where},m=n+p,\left(p=1,2,3,...\right)\mathrm{and}R\mathrm{is}\mathrm{Rydberg}\mathrm{constant}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For},p\mathit{}\mathit{}n\phantom{\rule{0ex}{0ex}}{v}_{mn}\mathit{=}cR{Z}^{\mathit{2}}\mathit{[}\frac{\mathit{1}}{{\mathit{n}}^{\mathit{2}}}{\left(\mathrm{1}+\frac{p}{n}\right)}^{\mathit{-}\mathit{2}}\mathit{-}\frac{\mathit{1}}{{\mathit{n}}^{\mathit{2}}}\mathit{]}\phantom{\rule{0ex}{0ex}}{v}_{mn}\mathit{=}cR{Z}^{\mathit{2}}\left[\frac{1}{{n}^{2}}-\frac{2p}{{n}^{3}}-\frac{1}{{n}^{2}}\right]\left[\mathrm{By}\mathrm{binomial}\mathrm{theorem}\right]\phantom{\rule{0ex}{0ex}}{v}_{mn}=cR{Z}^{\mathit{2}}\frac{2p}{{n}^{3}}\simeq \left(\frac{2cR{Z}^{2}}{{n}^{3}}\right)p$

Thus, the first few frequencies of light that is emitted when electrons fall to the* n*^{th} level from levels higher than *n* , are approximate harmonic (i.e., in the ratio 1 : 2 : 3 …) when *n* >> 1.

#### Page No 79:

#### Question 23:

What is the minimum energy that must be given to a H atom in ground state so that it can emit an *H*_{γ} line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such *H*_{γ} photon?

#### Answer:

${H}_{\gamma}$ in Balmer series corresponds to transition *n* = 5 to *n* = 2. So, the electron in ground state i.e., from *n* = 1must first be placed in state *n* = 5.

Energy required for the transition from *n* = 2 to *n* = 5 is given by

$\Rightarrow {E}_{2}-{E}_{1}=13.6-0.54=13.06\mathrm{eV}$

Since, angular momentum is conserved, angular momentum corresponding to H_{g} photon = change in angular momentum of electron

$\Rightarrow {L}_{2}-{L}_{1}=5h-2h=3h\phantom{\rule{0ex}{0ex}}\Rightarrow 3.18\times {10}^{-34}{\mathrm{kgm}}^{2}/\mathrm{s}$

#### Page No 79:

#### Question 24:

The first four spectral lines in the Lyman serics of a H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

#### Answer:

Given that the first four spectral lines in the Lyman series of a H-atom are:*λ*_{H} = 1218 Å, 1028Å, 974.3 Å and 951.4Å.

Reduced mass of H-atom,

${\mu}_{\mathrm{H}}=\frac{{m}_{e}}{1+{\displaystyle \frac{{m}_{e}}{M}}}={m}_{e}{\left(1+\frac{{m}_{e}}{M}\right)}^{-1}\simeq {m}_{e}\left(1-\frac{{m}_{e}}{M}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\mathrm{reduced}\mathrm{mass}\mathrm{of}\mathrm{D},\phantom{\rule{0ex}{0ex}}{\mu}_{\mathrm{H}}=\frac{{m}_{e}}{1+{\displaystyle \frac{{m}_{e}}{2M}}}={m}_{e}{\left(1+\frac{{m}_{e}}{2M}\right)}^{-1}\simeq {m}_{e}\left(1-\frac{{m}_{e}}{2M}\right)\phantom{\rule{0ex}{0ex}}$

If an electron jumps from higher orbit m to lower orbit n, then energy released:

${E}_{n}-{E}_{m}=h\nu =\frac{hc}{\lambda}\propto \mu \phantom{\rule{0ex}{0ex}}\Rightarrow \lambda \propto \frac{1}{\mu}$

So, the ratio,

$\frac{{\lambda}_{\mathrm{D}}}{{\lambda}_{\mathrm{H}}}=\frac{{\mu}_{\mathrm{H}}}{{\mu}_{\mathrm{D}}}=\frac{\left(1-{\displaystyle \frac{{m}_{e}}{M}}\right)}{\left(1-\frac{{m}_{e}}{2M}\right)}\phantom{\rule{0ex}{0ex}}=\left(1-\frac{{m}_{e}}{M}\right){\left(1-\frac{{m}_{e}}{2M}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\left(1-\frac{{m}_{e}}{M}\right)\left(1+\frac{{m}_{e}}{2M}\right)=\left(1+\frac{{m}_{e}}{2M}-\frac{{m}_{e}}{M}-\frac{{m}_{e}}{M}\times \frac{{m}_{e}}{2M}\right)\phantom{\rule{0ex}{0ex}}=\left(1-\frac{{m}_{e}}{2M}-\frac{{{m}_{e}}^{2}}{2{M}^{2}}\right)=\left(1-\frac{{m}_{e}}{2M}\right)\left({m}_{e}M\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\lambda}_{\mathrm{D}}}{{\lambda}_{\mathrm{H}}}=\left(1-\frac{{m}_{e}}{2M}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\mathrm{D}}=\left(1-\frac{{m}_{e}}{2M}\right){\lambda}_{\mathrm{H}}=\left(1-\frac{1}{2\times 1840}\right){\lambda}_{\mathrm{H}}\left(\because \frac{{m}_{e}}{M}=\frac{1}{1840}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\mathrm{D}}=0.99973{\lambda}_{\mathrm{H}}...\left(1\right)$

By putting the different values of wavelengths of spectrum released in Lyman series of H-atom $\left({\lambda}_{\mathrm{H}}\right)$ in equation (1),

we can get the corresponding values of ${\lambda}_{\mathrm{D}}$ = 1217.7 Å, 1027.7 Å, 974.04 Å and 951.143 Å.

#### Page No 79:

#### Question 25:

Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ^{1}H and ^{2}H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass *μ*, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here *μ* = *m*_{e}*M*/(*m _{e}* +

*M*) where

*M*is the nuclear mass and

*m*is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in

_{e}^{1}H and

^{2}H. (Mass of

^{1}H

nucleus is 1.6725 × 10

^{–27}kg, Mass of

^{2}H nucleus is 3.3374 × 10

^{–27}kg, Mass of electron = 9.109 × 10

^{–31}kg.)

#### Answer:

Let ${\mu}_{H}$ be the reduced mass of the mass of the Hydrogen and ${\mu}_{D}$ be the reduced mass of Deutarium.

Energy in the nth orbit is given by

${E}_{n}=\u2013\frac{\mu {z}^{2}{e}^{4}}{8{\epsilon}_{0}^{2}{h}^{2}}\left(\frac{1}{{n}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{for}\mathrm{Hydrogen}\phantom{\rule{0ex}{0ex}}\mathrm{So},{E}_{n}=\u2013\frac{{\mu}_{H}{z}^{2}{e}^{4}}{8{\epsilon}_{0}^{2}{h}^{2}}\left(\frac{1}{{n}^{2}}\right)\phantom{\rule{0ex}{0ex}}$

So, energy released in first Lyman series (i.e, *n* = 2 to *n* = 1)

${E}_{2\mathrm{to}1}=h{\nu}_{H}=\frac{{\mu}_{H}\times {1}^{2}\times {e}^{4}}{8{\epsilon}_{0}^{2}{h}^{2}}\left(\frac{1}{{1}^{2}}\u2013\frac{1}{{2}^{2}}\right)\left(\because z=1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{hc}{{\lambda}_{H}}=\frac{3}{4}\times \frac{{\mu}_{H}{e}^{4}}{8{\epsilon}_{0}^{2}{h}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{H}=\frac{4}{3}\times \frac{8{\epsilon}_{0}^{2}{h}^{3}c}{{\mu}_{H}{e}^{4}}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\phantom{\rule{0ex}{0ex}}{\lambda}_{D}=\frac{4}{3}\times \frac{8{\epsilon}_{0}^{2}{h}^{3}c}{{\mu}_{D}{e}^{4}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\u2206\lambda \%=\frac{{\lambda}_{D}-{\lambda}_{H}}{{\lambda}_{H}}\times 100\phantom{\rule{0ex}{0ex}}\mathrm{But},\frac{{\lambda}_{D}-{\lambda}_{H}}{{\lambda}_{H}}\times 100=\frac{{\mu}_{D}-{\mu}_{H}}{{\mu}_{H}}\times 100.....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{reduced}\mathrm{mass},\phantom{\rule{0ex}{0ex}}{\mu}_{H}=\frac{{m}_{e}{M}_{H}}{{m}_{e}+{M}_{H}}\left[\begin{array}{c}{M}_{H}=\mathrm{mass}\mathrm{of}\mathrm{Hydrogen}\\ {M}_{D}=\mathrm{mass}\mathrm{of}\mathrm{Deutrium}\\ {m}_{e}=\mathrm{mass}\mathrm{of}\mathrm{electron}\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{and}{\mu}_{D}=\frac{{m}_{e}{M}_{D}}{{m}_{e}+{M}_{D}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Putting this reduced mass in equation (1) we get.

$\frac{{\lambda}_{D}\u2013{\lambda}_{H}}{{\lambda}_{H}}\times 100=\left[\left(\frac{{m}_{e}+{M}_{H}}{{m}_{e}+{M}_{D}}\right)\frac{{M}_{D}}{{M}_{H}}\u20131\right]\times 100$

We know that:*m _{e}* <

*M*<

_{H}*M*

_{D}$\begin{array}{rcl}& \Rightarrow & \frac{\u2206\lambda}{{\lambda}_{H}}\times 100=\left[\frac{\left({\displaystyle \frac{{m}_{e}}{{M}_{H}}}+{\displaystyle \frac{{M}_{H}}{{M}_{H}}}\right)}{\left({\displaystyle \frac{{m}_{e}}{{M}_{D}}}+{\displaystyle \frac{{M}_{D}}{{M}_{D}}}\right)}\u20131\right]\times 100\\ & =& \left[\left(1+\frac{{m}_{e}}{{M}_{H}}\right){\left(1+\frac{{m}_{e}}{{M}_{D}}\right)}^{\u20131}\u20131\right]\times 100\\ & =& \left[\left(1+\frac{{m}_{e}}{{M}_{H}}\right)\left(1+\frac{{m}_{e}}{{M}_{D}}\right)\u20131\right]\times 100\end{array}$

Putting the required values and considering the assumption,

*m*<<

_{e}*M*

_{D}_{ }, We get:

The percentage difference in wavelength,

$=9.1\times {10}^{-31}\left[\frac{1}{3.3374\times {10}^{-27}}-\frac{1}{1.6725\times {10}^{-27}}\right]\times 100\phantom{\rule{0ex}{0ex}}=-2.7\times {10}^{-2}\%$

Here, negative sign shows that ${\lambda}_{D}<{\lambda}_{H}$.

#### Page No 79:

#### Question 26:

If a proton had a radius *R* and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) *R* = 0.1Å, and (ii) *R* = 10 Å.

#### Answer:

If an electron (*e*) revolves round the nucleus with speed *v *in radius *r*, the coulombic force provides the necessary centripetal force to revolve in a circular stationary orbit.

Angular monentum of the electron,

$m\nu r=\frac{h}{2\pi}.....\left(1\right)\left(\mathrm{In}\mathrm{ground}\mathrm{state}n=1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Also},\frac{m{\nu}^{2}}{r}=\frac{1}{4\pi {\epsilon}_{0}}\times \frac{{e}^{2}}{{r}^{2}}.....\left(2\right)$

Solving (1) and (2) we get:

$r=\frac{{h}^{2}}{4{\pi}^{2}}\times \frac{1}{m}\times \frac{4\pi {\epsilon}_{0}}{{e}^{2}}.....\left(3\right)$

Putting all the values we get:*r* = 0.53 Å

But potential energy of the electron in ground state.

$\begin{array}{rcl}{E}_{p}& =& -\frac{1}{4\pi {\epsilon}_{0}}\times \frac{{e}^{2}}{r}\\ & =& -\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{\u201319}\right)}^{2}}{0.53\times {10}^{\u201310}}J\\ & =& -\frac{9\times {10}^{9}\times \left(1.6\times {10}^{\u201319}\right)}{0.53\times {10}^{\u201310}\times 1.6\times {10}^{\u201319}}eV\\ & =& \u201327.2eV\end{array}$

Thus, kinetic energy,

$KE=\left|\frac{PE}{2}\right|=\left|\frac{\u201327.2}{2}\right|=+13.6eV$

Now, for shperical nucleus of radius R,

(i) For *R* = 0.1Å < *r* = 0.53Å

All the results calculated above will be same.

(ii) For *R* = 10Å > *r* = 0.53Å

i.e. electron moves inside the sphere with radius *r'*

$\mathrm{Charge}\mathrm{density}=\frac{e}{{\displaystyle \frac{4}{3}}\pi {R}^{3}}$

Charge within the radius *r'* will be

$=\left(\frac{e}{{\displaystyle \frac{4}{3}\pi {R}^{3}}}\right)\times \left(\frac{4}{3}\pi r{\text{'}}^{3}\right)=e\times {\left(\frac{r\text{'}}{R}\right)}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{radius},r\text{'}=\frac{{h}^{2}}{4{\pi}^{2}}\times \frac{1}{m}\times \frac{4\pi {\epsilon}_{0}}{{e}^{2}}\times {\left(\frac{R}{r\text{'}}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(0.53\mathrm{\AA}\right)\times {\left(\frac{R}{r\text{'}}\right)}^{3}\left[\mathrm{from}\mathrm{equation}\left(3\right)\right]\phantom{\rule{0ex}{0ex}}\left(\because \frac{{\mathrm{h}}^{2}}{4{\mathrm{\pi}}^{2}}\times \frac{1}{\mathrm{m}}\times \frac{4{\mathrm{\pi \epsilon}}_{0}}{\mathrm{m}}=0.53\mathrm{\AA}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow r\text{'}\times {\left(r\text{'}\right)}^{3}=\left(0.53\mathrm{\AA}\right)\times {R}^{3}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{\mathit{(}r\mathit{\text{'}}\mathit{)}}^{\mathit{4}}\mathit{=}\left(0.53\mathrm{\AA}\right)\mathit{\times}{R}^{\mathit{3}}$

It is given that *R* = 10Å

$\begin{array}{rcl}\mathrm{So},r\text{'}& =& {\left[0.53\times {10}^{3}\times {\left({10}^{\u201310}\right)}^{4}\right]}^{\frac{1}{4}}\\ & =& {\left(530\right)}^{\frac{1}{4}}\times {10}^{\u201310}\mathrm{m}\\ & =& {\left(530\right)}^{\frac{1}{4}}\mathrm{\AA}\end{array}$

So, kinetic energy,

$=\frac{1}{2}m{v}^{2}=\frac{m}{2}\times \frac{{h}^{2}}{4{\pi}^{2}\times {m}^{2}\times {\left({r}^{1}\right)}^{2}}$

Putting all the required values we get:

K.E. = 0.16 eV

Similarly:

Potential energy,

$\begin{array}{rcl}\mathrm{P}.\mathrm{E}.& =& \left(\frac{{e}^{2}}{4\pi {\epsilon}_{0}}\right)\left(\frac{{r\text{'}}^{2}-3{R}^{2}}{2{R}^{3}}\right)\\ & =& \left[\frac{{e}^{2}}{4\pi {\epsilon}_{0}}\times \frac{1}{r}\right]\times \left[\frac{r\left(r{\text{'}}^{2}\u20133{R}^{2}\right)}{2{R}^{3}}\right]\\ & =& \u20133.83\mathrm{eV}\end{array}$

Total energy of the electron = K.E. + P.E. = 0.16 eV −3.83 eV = −3.67 eV

#### Page No 79:

#### Question 27:

In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a *n *= 2 to *n* = 1 transition.

#### Answer:

We can neglect the recoil momentum of the massive nucleus and hence the total energy of the transition may be transferred to the Auger electron.

Number of valance electron in in Chromium = 1, so its energy can be given by Bohr's model as:

${E}_{n}=\u2013{Z}^{2}Rhc\frac{1}{{n}^{2}}$

Where, *R* is Rydberg constant,

*Z* is atomic number = 24 for Chromium* n *is orbit number

Now, if the electron jumps form second orbit (*n* = 2) to first orbit (*n* =1) then energy released in this process,

$\u2206E={Z}^{2}Rhc\left(\frac{1}{{1}^{2}}\u2013\frac{1}{{2}^{2}}\right)=\frac{3}{4}{Z}^{2}Rhc$

But, energy required to eject an electron from fourth orbit (*n* = 4) will be

${E}_{4}={Z}^{2}Rhc\frac{1}{{4}^{2}}=\frac{1}{16}{Z}^{2}Rhc$

Hence, the kinetic energy of the Auger electron,

$\begin{array}{rcl}K.E.& =& {Z}^{2}Rhc\left(\frac{3}{4}\right)-{Z}^{2}Rhc\left(\frac{1}{16}\right)\\ & =& {Z}^{2}Rhc\left(\frac{3}{4}-\frac{1}{16}\right)\\ & =& \frac{11}{16}{Z}^{2}Rhc\\ & =& \frac{11}{16}\times {\left(24\right)}^{2}\times 13.6\mathrm{eV}\\ & =& 5385.6\mathrm{eV}\end{array}$

This is the required kinetic energy in fourth orbit (*n* = 4).

#### Page No 80:

#### Question 28:

**F**|$=\frac{{e}^{2}}{\left(4\mathrm{\pi}{\epsilon}_{0}\right).{r}^{2}}$ force between an electron and a proton. The $\frac{1}{r}$ dependence of |

**F**|can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass

*m*, force would be modified to $\left|\mathbf{F}\right|=\frac{{e}^{22}}{\left(4\mathrm{\pi}{\epsilon}^{2}\right){r}^{2}}\left[\frac{1}{{r}^{2}}+\frac{\lambda}{r}\right].\mathrm{exp}\left(-\lambda r\right)\mathrm{where}\lambda ={m}_{p}c/\hslash \mathrm{and}\hslash =\frac{h}{2\mathrm{\pi}}$Estimate the change in the ground state energy of a H atom if

_{p}*m*were 10

_{p}^{–6}times the mass of an electron.

#### Answer:

If photon had a mass *m _{p}* then energy of the photon can be given as,

*E*=

_{p}*m*

_{p}*c*

^{2}(∴

*c*= speed of light)

(It is given that

*m*= 10

_{p}^{–6}×

*m*)

_{e}*E*

_{p}_{}= 10

^{–6}×

*m*×

_{e}*c*

^{2}

= 10

^{–6}× 9.1 × 10

^{–31}× (3 × 10

^{8})

^{2}J

⇒

*m*

_{p}c^{2 }≈ 0.8 × 10

^{–19 }J

But we know that,

Angular momentum,

${m}_{p}\times c\times r=\frac{h}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow r=\frac{h}{2\pi}\times \frac{1}{{m}_{p}\times c}=\frac{hc}{2\pi}\times \frac{1}{\left({m}_{p}{c}^{2}\right)}.....\left(1\right)$

Putting all the values we get

⇒

*r*≈ 4 × 10

^{–7}m

This value of radius is much more than the Bohr's radius.

Now modified force,

$F=\frac{{e}^{2}}{4\mathrm{\pi}{\epsilon}_{0}}\left[\frac{1}{{r}^{2}}+\frac{\lambda}{r}\right]{e}^{\u2013\lambda r}\phantom{\rule{0ex}{0ex}}\mathrm{Here},\lambda =\frac{{m}_{p}c}{h}\times 2\mathrm{\pi}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda}=\frac{h}{2\mathrm{\pi}{m}_{p}c}\phantom{\rule{0ex}{0ex}}\frac{1}{\lambda}\approx 4\times {10}^{\u20137}{\mathrm{m}}^{\u20131}\left[\because =\frac{h}{2\mathrm{\pi}{m}_{p}c}\approx 4\times {10}^{\u20137}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda}{r}_{\mathrm{Bohr}\text{'}\mathrm{s}\mathrm{radius}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda {r}_{\mathit{B}}1\phantom{\rule{0ex}{0ex}}$

Now , potential energy, $U\left(r\right)=-\frac{{e}^{2}}{4\pi {\epsilon}_{0}}\times \frac{{e}^{-\lambda r}}{r}.....\left(3\right)$

From, Bohr's model,

$mvr\text{'}=\frac{h}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{h}{2\mathrm{\pi}mr}$

$\mathrm{also},\frac{m{v}^{2}}{r}=\frac{{e}^{2}}{4\mathrm{\pi}{\epsilon}_{0}}\left[\frac{1}{{r}^{2}}+\frac{\lambda}{r}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{h}^{2}}{4{\mathrm{\pi}}^{2}m}=\left(\frac{{e}^{2}}{4\mathrm{\pi}{\epsilon}_{0}}\right)\left(r+\lambda r\right)$

But if,

*λ*= 0

$r={r}_{\mathrm{B}}=\frac{h}{2\mathrm{\pi}m}\times \frac{4\mathrm{\pi}{\epsilon}_{0}}{{e}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{h}^{2}}{4{\mathrm{\pi}}^{2}m}=\frac{{e}^{2}}{4\mathrm{\pi}{\epsilon}_{0}}\times {r}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\mathrm{As},{\lambda}^{-1}{r}_{\mathrm{B}}$

Let,

*r*=

*r*

_{B }+ δ

$\Rightarrow 0=\lambda {r}_{\mathrm{B}}^{2}+\delta \left(1+2\lambda {r}_{\mathrm{B}}\right)\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathrm{Neglecting}\mathrm{the}\mathrm{heigher}\mathrm{power}\mathrm{of}\mathrm{\delta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \delta \mathit{=}\frac{\mathit{-}\lambda {r}_{\mathrm{B}}^{\mathit{2}}}{1+2\lambda {r}_{\mathrm{B}}}\mathit{\approx}\lambda {r}_{\mathrm{B}}^{\mathit{2}}\mathit{}\mathit{(}1\mathit{-}\mathit{2}\lambda {r}_{B}\mathit{)}\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=-\lambda {r}_{\mathrm{B}}^{2}\left[\because \lambda {r}_{\mathrm{B}}1\right]$

Now potential energy,

$\mathrm{P}.\mathrm{E}.=-\frac{{e}^{2}\times {e}^{-\lambda \delta -\lambda {r}_{\mathrm{B}}}}{4\mathrm{\pi}{\epsilon}_{0}\times \left({r}_{\mathrm{B}}+\delta \right)}\phantom{\rule{0ex}{0ex}}\approx -27.2\mathrm{eV}\left(\mathrm{remains}\mathrm{constant}\right)$

Now, kinetic energy,

$\begin{array}{rcl}\mathrm{K}.\mathrm{E}.& =& -\frac{1}{2}m{v}^{2}=-\frac{1}{2}m\times {\left(\frac{h}{2\mathrm{\pi}}\right)}^{2}\times \frac{1}{m{r}^{2}}\\ & =& \frac{{h}^{2}}{4{\mathrm{\pi}}^{2}}\times \frac{1}{2{\left({r}_{\mathrm{B}}+\delta \right)}^{2}}\\ & =& \frac{{h}^{2}}{4{\mathrm{\pi}}^{2}}\times \frac{1}{2{r}_{\mathrm{B}}^{2}}\times \left(1-\frac{2\delta}{{r}_{\mathrm{B}}}\right)\left[\because \delta {r}_{\mathrm{B}}\right]\\ & =& \left(13.6\mathrm{eV}\right)\left[1+2\lambda {r}_{\mathrm{B}}\right]\end{array}$

Thus, total energy,

= P.E. + K.E.

= – 27.2 eV + 13.6[1 + 2

*λr*

_{B}] eV

= [–13.6 + 27.2

*λr*

_{B}] eV

Hence, net change in energy = [–13.6 + 27.2

*λr*

_{B}– (–13.6)] eV

*λr*

_{B}eV

*λr*

_{B}eV.

#### Page No 80:

#### Question 29:

The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge + *q*_{1}, –*q*_{2} is modified to

$\left|\mathbf{F}\right|=\frac{{q}_{1}{q}_{2}}{\left(4{\mathrm{\pi \epsilon}}_{0}\right){r}^{2}},r\ge {R}_{0}\phantom{\rule{0ex}{0ex}}=\frac{{q}_{1}{q}_{2}}{4{\mathrm{\pi \epsilon}}_{0}}\frac{1}{{{R}_{0}}^{2}}{\left(\frac{{R}_{0}}{r}\right)}^{\epsilon}.r\le {R}_{0}$

Calculate in such a case, the ground state energy of a H-atom, if ε = 0.1, *R*_{0} = 1Å.

#### Answer:

Taking the case when,* r* ≤ *R*_{o} = 1 Å

Now, coulombic force between the charge *q*_{1} and *q*_{2},

$\begin{array}{rcl}F& =& \frac{{q}_{1}{q}_{2}}{4\mathrm{\pi}{\epsilon}_{\mathrm{o}}{R}_{\mathrm{o}}^{2}}\times {\left(\frac{{R}_{\mathrm{o}}}{r}\right)}^{\epsilon}\\ & =& \frac{{q}_{1}{q}_{2}}{4\mathrm{\pi}{\epsilon}_{\mathrm{o}}}\times \frac{{R}_{\mathrm{o}}^{\left(\epsilon -2\right)}}{{r}^{\epsilon}}\end{array}$

Now, let *ε* =* δ* + 2

$\mathrm{So},F=\frac{{q}_{1}{q}_{2}}{4\mathrm{\pi}{\epsilon}_{o}}\times \frac{{R}_{\mathrm{o}}^{\delta}}{{r}^{\delta +2}}\phantom{\rule{0ex}{0ex}}\mathrm{If}\left|{q}_{1}\right|=\left|{q}_{2}\right|=1.6\times {10}^{-19}\mathrm{C}$

Then,

$F=9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}\times \frac{{R}_{\mathrm{o}}^{\delta}}{{r}^{\delta +2}}\phantom{\rule{0ex}{0ex}}=\left(23.04\times {10}^{-29}\right)\frac{{R}_{\mathrm{o}}^{\delta}}{{r}^{\delta +2}}\phantom{\rule{0ex}{0ex}}\mathrm{Let},\Lambda =23.04\times {10}^{-29}\phantom{\rule{0ex}{0ex}}\mathrm{Then}F=\Lambda \frac{{R}_{\mathrm{o}}^{\delta}}{{r}^{\delta +2}}$

This coulombic force provides necessary centripetal force to revolve in circular orbit.

$\mathrm{So},\frac{m{v}^{2}}{r}=\Lambda \left(\frac{{R}_{\mathrm{o}}^{\delta}}{{r}^{\delta +2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}=\Lambda \frac{{R}_{\mathrm{o}}^{\delta}}{m{r}^{\delta +2}}\times r=\Lambda \frac{{R}_{\mathrm{o}}^{\delta}}{m{r}^{\delta +1}}$

But angular momentum of the electron revolving in *n*^{th} Bohr's orbit,

$mvr=n\frac{h}{2\mathrm{\pi}}=n\hslash \left(\mathrm{Here},\hslash =\frac{h}{2\mathrm{\pi}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{nh}{2\mathrm{\pi}mv}=\frac{nh}{2\mathrm{\pi}m}\times {\left(\frac{m{r}^{1+\delta}}{\Lambda {R}_{\mathrm{o}}^{\delta}}\right)}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Radius}\mathrm{of}{n}^{\mathrm{th}}\mathrm{orbit},{r}_{n}={\left[\frac{{n}^{2}{\hslash}^{2}}{m\Lambda {R}_{\mathrm{o}}^{\delta}}\right]}^{\frac{1}{1-\delta}}$

Now for *n* = 1 (first orbit)

${r}_{1}={\left[\frac{{1}^{2}\times {\hslash}^{2}}{m\Lambda {R}_{\mathrm{o}}^{\delta}}\right]}^{\frac{1}{1-\delta}}$

Now putting the required values, we get:

${r}_{1}={\left[\frac{1.05\times {10}^{-68}}{9.1\times {10}^{-31}\times 2.3\times {10}^{-28}\times {10}^{19}}\right]}^{\left(\frac{1}{2.9}\right)}\phantom{\rule{0ex}{0ex}}\left(\begin{array}{l}\because \epsilon =\delta +2\\ \Rightarrow \delta =\epsilon -2\\ =0.1-2=-1.9\\ \Rightarrow 1-\delta =1+1.9=2.9\end{array}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},{r}_{1}\approx 8\times {10}^{-11}\mathrm{m}\phantom{\rule{0ex}{0ex}}{r}_{1}\approx 0.08\mathrm{nm}$

(ii) Velocity of the electron in *n*^{th} orbit,

${v}_{n}=n\frac{h}{2\mathrm{\pi}m{r}_{n}}=n\frac{h}{2\mathrm{\pi}}\times {\left(\frac{m\Lambda {R}_{\mathrm{o}}^{\delta}}{{n}^{2}{\hslash}^{2}}\right)}^{\frac{1}{1-\delta}}$

For first orbit, *n* = 1

${v}_{1}=\frac{\hslash}{m{r}_{1}}\approx 1.44\times {10}^{6}\mathrm{m}/\mathrm{s}$

(iii) Kinetic energy of the electronic first orbit,

$\begin{array}{rcl}\mathrm{K}.\mathrm{E}.& =& \frac{1}{2}m{v}_{1}^{2}\approx 9.43\times {10}^{-19}\mathrm{J}\\ & =& \frac{9.43\times {10}^{-19}}{1.6\times {10}^{-19}}\mathrm{eV}\approx 5.9\mathrm{eV}\end{array}$

Now, potential energy upto *R*_{o},

$\mathrm{P}.\mathrm{E}{.}_{1}=-\frac{\Lambda}{{R}_{\mathrm{o}}}$

P.E. from *R*_{o }to *r*,

$\begin{array}{rcl}\mathrm{P}.\mathrm{E}{.}_{2}& =& \Lambda {R}_{\mathrm{o}}^{\delta}{\int}_{{R}_{\mathrm{o}}}^{r}\frac{dr}{{r}^{2}+\delta}\\ & =& \frac{\Lambda {R}_{\mathrm{o}}^{\delta}}{1-\left(2+\delta \right)}{\left[\frac{1}{{r}^{1+\delta}}\right]}_{{R}_{\mathrm{o}}}^{r}\\ & =& -\frac{\Lambda {R}_{\mathrm{o}}^{\delta}}{1+\delta}\left[\frac{1}{{r}^{1+\delta}}-\frac{1}{{R}_{\mathrm{o}}^{1+\delta}}\right]\end{array}$

So, total potential energy

$\begin{array}{rcl}\mathrm{P}.\mathrm{E}{.}_{1}+\mathrm{P}.\mathrm{E}{.}_{2}& =& -\frac{\Lambda}{{R}_{\mathrm{o}}}-\frac{\Lambda {R}_{o}^{\delta}}{1+\delta}\left[\frac{1}{{r}^{1+\delta}}-\frac{1}{{R}_{\mathrm{o}}^{1+\delta}}\right]\\ & =& -\frac{\Lambda}{1+\delta}\left[\frac{{R}_{\mathrm{o}}^{\delta}}{{r}^{1+\delta}}-\frac{1}{{R}_{\mathrm{o}}}+\frac{1+\delta}{{R}_{\mathrm{o}}}\right]\\ & =& -\frac{\Lambda}{1+\left(-1.9\right)}\times \left[\frac{{R}_{\mathrm{o}}^{-1.9}}{{r}^{1+\left(-1.9\right)}}-\frac{1.9}{{R}_{\mathrm{o}}}\right]\end{array}$

$\Rightarrow \mathrm{P}.\mathrm{E}.=\frac{2.3}{0.9}\times {10}^{-18}\left[{\left(0.8\right)}^{0.9}-1.9\right]\mathrm{J}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}.\mathrm{E}.=\frac{1}{1.6\times {10}^{-19}}\times \frac{2.3}{0.9}\times {10}^{-19}\left[{\left(0.8\right)}^{0.9}-1.9\right]\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}.\mathrm{E}.=-17.3\mathrm{eV}$

Hence, the total energy of the electron,

T.E. = P.E. + K.E.

â€‹= – 11.4 eV

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