NCERT Exemplar Class 12 Physics Solutions Chapter 12 Atoms

Question:2

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is
$B=-\frac{me^{2}}{8n^{2}\varepsilon _{0}^{2}h^{2}}$ . (m = electron mass). If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be

$B=-\frac{me^{2}}{8n^{2}\varepsilon _{0}^{2}h^{2}}$ (M = proton mass)
This last expression is not correct because
A. n would not be integral.
B. Bohr-quantization applies only to electrons
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:

The answer is option (C), the frame in which the electron is at rest is not inertial.

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 above-mentioned Bohr model is not relevant for calculating energy levels of atoms containing multiple electrons because of the assumption that the centripetal force is provided by the nucleus’s electrostatic force of attraction, and it is also used as an assumption while deriving formulas for radius/energy levels, etc. Therefore, the model is only applicable to single-electron atoms. If applied to multiple electron atoms, another factor, repulsion due to the electrons, needs to be accounted for. This proves that the simple Bohr model cannot be used to calculate the energy level of atoms with numerous electrons.

The answer is option (A).

Question:4

For the ground state, the electron in the H-atom has an angular momentum =? 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:

Option (A) is the correct answer as -
Neil Bohr proposed that the magnitude of electron’s angular momentum is quantized
$i.e\; L=mv_{n}r_{n}=n\left ( \frac{h}{2\pi} \right )where\; n=1,2,3.........$
each value of n corresponds to a permitted value of the orbit radius.
r_n=Radius of n^th, v_n=corresponding speed
Angular momentum given by Bohr’s model is a vector quantity, and the model only gives the magnitude of the angular momentum. Therefore, angular momentum is not described to a full extent by Bohr’s model. So, the values given for the angular momentum of a revolving electron by Bohr’s model are not correct.

Question:5

O₂ 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:

The answer is option (A).
Nuclear forces are responsible for keeping the nucleons bound in the nucleus.
These forces come under short-range forces, and the maximum distance of their existence is from $10^{-15}$ m. These are one of the strongest forces present in nature. They bring stability to the nucleus, and they are attractive in nature. They are non-central and do not have any charge dependency.
Due to their strong nature, nuclear forces dominate the force between protons in the nucleus, also known as the repulsive Coulomb force. And as the distance increases more than a few femtometres, the nuclear forces reduce to zero between two nucleons.
And in case of the O₂ molecule made up of two oxygen atom molecules, the nuclear forces are not accounted for as they work within the nucleus, so (a) is the correct option.

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 answer is option (A).
Total energy: Total energy (E) is the sum of both potential energy and kinetic energy, i.e. E = K + U
(a) 10.20 eV
The lowest state of the atom, called the ground state is that of the lowest energy. The energy of this state (n=1), E₁ is -13.6 eV
Let us take two hydrogen atoms in the ground state. When two atoms collide inelastically, the total energy associated with the H atoms –
=2 X (13.6 eV)=27.2 eV
The maximum amount of energy by which their combined kinetic energy is reduced when any one of them goes into a first excited state (n = 2) after the inelastic collision.
Thus, the total energy associated with the two H atoms after the collision is -
$=\left ( \frac{13.6}{2^{2}} \right )+(13.6)=17.0 eV$
Hence, the maximum loss of their combined kinetic energy
=27.2-17.0=10.2 eV
Therefore, option (A) 10.20 eV is the right option.

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.

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 form of radiation.
C. because of momentum conservation.
D. because of angular momentum conservation.

Answer:

The correct answers are the options (A, B). Due to the law of conservation of energy, a moving electron and a proton cannot collectively form a hydrogen atom, as they will simultaneously release energy in the form of radiation.

Question:10

The Bohr model for the spectra of a H-atom
A. will not be applicable to hydrogen in the molecular form.
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 correct answers are the options (A, B). The model of the atom proposed by Neil Bohr was applicable for the hydrogen atom and some other light atoms containing only one electron around the nucleus, which is stationary having a positive charge, Ze (called a hydrogen-like atom, e.g., H, $He^{+}$, $Li^{+2}$, $Na^{+1}$ etc). But one of the limitations of the model is that it is not applicable to hydrogen in its molecular form, and also in the case of He atom.

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:

The correct answers are the options (B, D) as the correct options.
To observe the Balmer series for the H-atom, simply by measuring the frequencies of light that are emitted because of consequences such as the transition between higher excited states and the first excited state and as a sequence of frequencies with the higher frequencies getting closely packed.

Question:12

Let $E_{n}=\frac{-1}{8\varepsilon_{0}^{2}}\frac{me^{2}}{n^{2}h^{2}}$ be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E2-E1)/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:

The correct answers are the options (B, D). Let us assume $E_{1}$ and $E_{2}$ as the energy corresponding to n = 2 and n = 1 respectively. According to Bohr’s model of the atom, few of the atoms will reach the first excited state if the radiation of energy on a sample in which all the hydrogen atoms are at the ground state is $\Delta$E = ($E_2 – E_1$) = hf incident. But as the energy is insufficient for the transition to take place from n = 1 to n =3, none of the atoms present will reach up to the n = 3 state.

Question:13

The simple Bohr model 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 correct option of the mentioned problem is (C, D) as,
The famous model of atomic structure proposed by Neil Bohr is only applicable to light atoms j = having only one electron revolving around the nucleus, and it is mostly used for the H-atom. The electron must revolve around a nucleus which stationary and positively charged. It is also applicable for - H, $He^{+}$, $Li^{+2}$, $Na^{+1}$ etc.
Because of has only one electron and electrons are not subjected to centripetal forces; thus, it is also applicable for He⁴.

Question:15

Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of the Bohr model, will be very close. Explain why.

Answer:

One of the biggest limitations of Bohr’s model of the atom is that it is only applicable for either light atoms having one electron revolving around the nucleus or for H-atoms. The nucleus of these atoms must be positively charged and stationary; such atoms are termed Ze (called a hydrogen-like atom). But if one electron is removed from both He⁴ and He³, they will fulfil the condition, and the model can be used to find their energy levels.

Question:16

When an electron falls from 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:

Electrons are negatively charged particles, and when they undergo a transition from a higher energy level to a lower energy level, they are accelerated. Thus being accelerated particles, they emit energy in the form of electromagnetic radiations.

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:

Neil Bohr proposed that if an electron is moving around a stationary nucleus, the required centripetal force is provided by the electrostatic force of attraction.
i.e $\frac{1}{4 \pi \epsilon _{0}}\frac{q_{1}q_{2}}{r^{2}}=\frac{mv^{2}}{r} ...............(i)$
Therefore, the magnitude of the electrostatic force $F \propto q_1 \times q_2$. The Bohr’s formula for the H-atom remains unchanged if the charge on the proton is (+4/3)e, whereas the charge carried by the electron is (-3/4)e. This is possible as Bohr’s formula requires the product of the charges, which remain constant for given values of charges.

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 the Bohr model, electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as
$L=\frac{nh}{2\pi}\;or\;L\propto n$

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 z=2 and for ground n=1
Thus ground state energy of a He atom
$E_{n}=-13.6\frac{Z^{2}}{n^{2}}eV=-13.6\frac{2^{2}}{I^{2}}eV=-54.4eV$
Thus, the ground state will have two electrons each of energy ,R and the total ground state energy would be -(4 X 13.6)eV=54.4eV

Question:21

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

Answer:

The equivalent electric current due to the rotation of charge is given by
$i=\frac{Q}{T}=Q \times f $ where f is the frequency
In a hydrogen atom, an electron in the ground state revolves on a circular orbit whose radius is equal to the Bohr radius. Let the velocity of the electron be v
Number of revolutions per unit time
$f=\frac{2\pi a_{0}}{v}$
The electric current is given by
$i=e\left ( \frac{2 \pi a_{0}}{v} \right )=\frac{2 \pi a_{0}}{v}e$

Question:22

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

Answer:

Frequency of emitted radiation:
$\Delta E=hf\Rightarrow f=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}=RcZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )$
Wavenumber:
Wavenumber is the number of waves in a unit length
$\bar{v}=\frac{1}{\lambda}=\frac{f}{c}\Rightarrow \frac{1}{\lambda}=RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )=\frac{13.6Z^{2}}{hc}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )$
The number of spectral lines: If an electron jumps from a higher energy orbit to a lower energy orbit, it emits radiation with various spectral lines.
If an electron falls from n2 to n1, then the number of spectral lines emitted is given by
$N_{E}=\frac{\left ( n_{2}-n_{1}+1 \right )\left ( n_{2}-n_{1} \right )}{2}$
If an electron falls from the n^th orbit to ground state (i.e n₂=n and n₁=1), then the number of spectral lines emitted $N_{E}=\frac{\left ( n(n-1) \right )}{2}$
The frequency of any line in a series in the spectrum of hydrogen-like atoms corresponding to the transition of an electron from (n + p) level to the nth level can be expressed as a difference of two terms.
$f_{min}=cRZ^{2}\left [ \frac{1}{\left ( n+p \right )^{2}}-\frac{1}{n^{2}} \right ]$
where m=n+p,(p=1,2,3,.....) and R is Rydberg constant for p<<n

$\begin{aligned}
& f_{\min }=c R Z^2\left[\frac{1}{n^2}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^2}\right] \\
& f_{\min }=c R Z^2\left[\frac{1}{n^2}-\frac{2 p}{n^3}-\frac{1}{n^2}\right]
\end{aligned}$

$\left[\right.$ by Binomial theorem $\left.(1+x)^2=1+n x \quad I f|x|<1\right]$

$f_{\min }=c R Z^2 \frac{2 p}{n^3}=\left(\frac{2 c R Z^2}{n^3}\right) p$

Hence, the first few frequencies of light that are emitted when the electrons fall to the nth level from levels higher than n, are approximately harmonic (i.e., in ratio 1:2:3 …) when n>> 1.

Question:23

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

Answer:

“H-alpha” corresponds to transition 3 -> 2. And “H-beta” corresponds to 4 -> 2 and the “H-gamma” corresponds to transition 5 -> 2.
Where H is the element hydrogen, four of the Balmer lines are in the technically visible part of the spectrum, with wavelengths larger than 400nm and shorter than 700 nm.
From the above discussion, it is clear that H-gamma in the Balmer series corresponds to the transition n = 5 to n = 2. Initially, the electron in the ground state must be placed in state n = 5.

Energy required for the transition from n=1 to n=5 is given by
E=E₅-E₁
=$\left ( \frac{-13.6}{5^{2}} \right )-\left ( \frac{-13.6}{1^{2}} \right )=13.06 eV$
As angular momentum should be conserved, the angular momentum corresponding to H_y photon = the change in angular momentum of the electron
$=5\left ( \frac{5}{2\pi} \right )-2\left ( \frac{h}{2\pi} \right )=\frac{3h}{2\pi}\\ =\frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.14}=3.17 \times 10^{-34}Js$

Question:25

Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ${ }^1 \mathrm{H}$ and 2 H . This is because the wavelength of transition depends to a certain extent on the nuclear mass. If the 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 $\mu$, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\mu=m_e M /\left(m_e+M\right)$ where $M$ is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the $1^{\text {st }}$ line of the Lyman series in ${ }^1 \mathrm{H}$ and ${ }^2 \mathrm{H}$. (Mass of 1 H nucleus is $1.6725 \times 10^{-27} \mathrm{~kg}$, the mass of 2 H nucleus is $3.3374 \times 10^{-27} \mathrm{~kg}$, the mass of electron $=9.109 \times 10^{-31} \mathrm{~kg}$.)

Answer:

According to Bohr’s theory for a hydrogen-like atom of atomic number Z, the total energy of the electron in the nth state is
$E_{n}=-\left \{ \frac{\mu}{2h^{2}}\left (\frac{e^{2}}{4\pi \epsilon _{0}} \right )^{2} \right \}\frac{Z^{2}}{n^{2}}=\frac{\mu Z^{2}e^{4}}{8 \epsilon_{0}^{2}h^{2} }\left ( \frac{1}{n^{2}} \right )$
Where signs are usual, and the u that occurs in Bohr’s formula is the reduced mass of the electron and proton.
For a hydrogen atom:
$\mu_{H}=\frac{m_{e}M_{H}}{\left ( m_{e}+M_{H} \right )}$
For Deutrium
:$\mu_{D}=\frac{m_{e}M_{D}}{\left ( m_{e}+M_{D} \right )}$
Let $\mu_{H}$ be the reduced mass of hydrogen and $\mu_{D}$ that of Deuterium. Then the frequency of the 1st Lyman line in hydrogen is
$hf_{H}=\frac{\mu_{H}e^{4}}{8\epsilon _{0}^{2}h^{2}}\left ( 1-\frac{1}{4} \right )=\frac{\mu_{H}e^{4}}{8\epsilon _{0}^{2}h^{2}}\times \frac{3}{4}$
Thus, the wavelength of the transition is
$\lambda_{H}=\frac{3}{4}\frac{\mu_{H}e^{4}}{8\epsilon _{0}^{2}h^{2}c}$
The wavelength of the transition for the same line in detutrium is
$\lambda_{D}=\frac{3}{4}\frac{\mu_{D}e^{4}}{8\epsilon _{0}^{2}h^{2}c}$
$\therefore$ Difference in wavelength $\Delta \lambda=\lambda_{D}-\lambda_{H}$
Hence, the percentage difference is

$\begin{aligned}
& 100 \times \frac{\Delta \lambda}{\lambda_H}=\frac{\lambda_D-\lambda_H}{\lambda_H} \times 100=\frac{\mu_{D-\mu_H}}{\mu_H} \times 100 \\
= & \frac{\frac{m_e M_D}{\left(m_e+M_D\right)}-\frac{m_e M_H}{\left(m_e+M_H\right)}}{\frac{m_e M_H}{\left(m_e+M_H\right)}} \times 100 \\
= & {\left[\left(\frac{m_e+M_H}{m_e+M_D}\right) \frac{M_D}{M_H}-1\right] \times 100 }
\end{aligned}$

Since $m_e \ll M_H \ll M_D$

$\begin{aligned}
& \frac{\Delta \lambda}{\lambda_H} \times\left[\frac{M_H}{M_D} \times \frac{M_D}{M_H}\left(\frac{1+\frac{m_e}{M_H}}{1+\frac{m_e}{M_D}}\right)-1\right] \times 100 \\
= & {\left[\left(1+\frac{m_e}{M_H}\right)\left(1+\frac{m_e}{M_D}\right)^{-1}-1\right] \times 100 } \\
= & {\left[\left(1+\frac{m_e}{M_H}\right)\left(1+\frac{m_e}{M_D}\right)-1\right] \times 100 }
\end{aligned}$

$\left(\right.$ By binomial theorem $(1+x)^n=1+n x$ is $\left.|x|<1\right)$

$\frac{\Delta \lambda}{\lambda_H} \times 100=\left[1+\frac{m_e}{M_H}-\frac{m_e}{M_D}-\frac{\left(m_e\right)^2}{M_H M_D}-1\right] \times 100$

Neglecting $\frac{\left(m_e\right)^2}{M_H M_D}$ as it is very small

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:

The electron present in the H-atom revolves around the point-sized proton of a defined radius rB (Bohr’s radius) in a circular path when the atom is in the ground state.
$\begin{aligned}
& \text{As, } m v r_B=h \text { and } \frac{M v^2}{r_B}=\frac{-1 \times e \times e}{4 \pi \varepsilon_0 r_B^2} \\
& \frac{m}{r_B}\left(\frac{-h^2}{m^2 r_B^2}\right)=\frac{e^2}{4 \pi \epsilon_0 r_B^2} \\
& r_B=\frac{4 \pi \epsilon_0}{e^2} \frac{h^2}{m}=0.53 A^o \\
& K . E=\frac{1}{2} m v^2=\left(\frac{m}{2}\right)\left(\frac{h}{m r_B}\right)^2 =\frac{h^2}{2 m r_B^2}=13.6 \mathrm{eV}
\end{aligned}$
total energy of the electron, i.e
E=K+U=+13.6eV-27.2eV=-13.6eV
(i)when r=0.1Å:R<r_B (as r_B =0.51 Å )and the ground state energy is the same as obtained earlier for point size proton, i.e -13.6 eV
(ii)when r=10Å:R>>r_B the electrons move inside the proton (assumed to be a sphere of radius R ) with the new Bohr's radius r'_B
clearly
$r'_{B}=\frac{4 \pi \epsilon_{0}}{m(e)(e')}$
[Replacing e² by (e)(e') where e' is the charge on the sphere of radius r'_B ]
Since
$\begin{aligned}
e^{\prime}&=\left[\frac{e}{(4 \pi / 3) R^3}\right]\left[\left(\frac{4 \pi}{3}\right) r_B^{3 \prime}\right]=\frac{r_B^{3 \prime}}{R^3} \\
r_B^{3 \prime}&=\frac{4 \pi \epsilon_0 h^2}{m_e\left(e r_B^{3 \prime} R^3\right)}=\left(\frac{4 \pi \epsilon_0 h^2}{m_e^2}\right)\left(\frac{R^3}{r_B^{3 \prime}}\right) \\
\text {Or } r_B^{3 \prime}&=\left(\frac{4 \pi \epsilon_0 h^2}{m e^2}\right) R^3 \\
& =(0.51 A^o)(10 A^o)=510 A^o
\end{aligned}$

r'_B =4.8Å which is less than R(=10Å)
KE of the electron
$K^{\prime}=\frac{1}{2} m v^2=\left(\frac{m}{2}\right)\left(\frac{h^2}{m^2 r_B^{2 \prime}}\right)=\frac{h}{m r_B^{2 \prime}} =\left(\frac{h^2}{m r_B^{2 \prime}}\right)\left(\frac{r_B}{r_B^{\prime}}\right)=13.6 \mathrm{eV}\left(\frac{0.51 A^o}{4.8 A^o}\right)=0.16 \mathrm{eV}$
The potential at a point inside the charged proton,
i.e
$V=\frac{k_{e}}{R}\left (3-\frac{{r_{B}^{2}}'}{R^{2}} \right )=k_{e}e\left ( \frac{3R^{2}-{r_{B}^{2}}'}{R^{3}} \right )$

The potential energy of an electron and a proton,
$\begin{aligned}
\mathrm{u}=&-\mathrm{eV}\\
= & -e\left(k_e e\right)\left[\frac{3 R^2-r_B^2}{R^3}\right] \\
= & -\left(\frac{e^2}{4 \pi \varepsilon_0 r_B}\right)\left[\frac{r_B\left(3 R^2-r_B^2\right)}{R^3}\right] \\
= & -(24.2 e \mathrm{~V})\left[\frac{(0.51 A^o)(300 A^o-23.03 A^o)}{1000 A^o}\right] \\
= & 3.83 \mathrm{eV}
\end{aligned}$
Total energy of the electron
E=K+U=0.16eV-3.83eV=-3.67e

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:

Auger Effect: While responding to the downward transition by another electron in the atom, the characteristic energies of the atoms are ejected, and this process is called the Auger effect. The bombardment with high-energy electrons results in the formation of a vacancy in Auger spectroscopy. But the effect is possible when another interaction leads to vacancy formation. It is an electron arrangement method after the nucleus captures the electron.
During the removal of an electron shell from the atom, a simultaneous response by a higher-level electron will result in a downward transition and fill the vacancy. Multiple times, there is a release of photons along with the vacancy filling; the energy of these photons is equal to the energy gap between the upper and lower levels. This is also known as X-ray fluorescence due to the presence of all these heavy atoms in the X-ray region. But the same emission process for lighter atoms or outer electrons forms a line spectrum.
But in most cases, a higher-level electron fills the vacancy and emits a photon.

Because the nucleus is large and thus the momentum of the atom is neglected, and the complete transition energy is transferred to the Auger electron. The energy states represented by Bohr’s model are similar to the states of Cr due to the presence of a single valence electron in its atom.
The energy of the n^th state $E_{n}=-Rch\frac{Z^{2}}{n^{2}}=-13.6\frac{Z^{2}}{n^{2}}eV$ where R is the Rydberg constant and Z=24

The energy released in a transition from 2 to 1 is
$\Delta E=13.6Z^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )=13.6Z^{2}\left ( \frac{1}{1^{2}}-\frac{1}{2^{2}} \right )$
$=13.6Z^{2}\left ( 1-\frac{1}{4} \right )=13.6Z^{2}\times \frac{3}{4}$
The energy required of the Auger electron is
$KE=13.6Z^{2}\left ( \frac{3}{4}-\frac{1}{16} \right )=5385.6eV$

Question:28

The inverse square law in electrostatics is $\left | F \right |=\frac{e^{2}}{\left (4 \pi \epsilon _{0} \right )r^{2}}$ for the 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 mp, force would be modified to $\left | F \right |=\frac{e^{2}}{\left (4 \pi \epsilon _{0} \right )r^{2}}\left [ \frac{1}{r^{2}} +\frac{\lambda}{r}\right ].\exp \left ( -\lambda r \right )$ where $\lambda=m_{p}c/h$ and $h=\frac{h}{2\pi}$. Estimate the change in the ground state energy of a H-atom if mp were 10–6 times the mass of an electron.

Answer:

We are given
$\begin{aligned}
& \lambda=\frac{m_p c}{h}=\frac{m_p c^2}{h c}=\frac{\left(10^2 m_e\right) c^2}{h c} \\
&= \frac{10^{-6}[0.51]\left[1.6 \times 10^{-13} \mathrm{~J}\right] 3 \times 10^8 \mathrm{~ms}^{-1}}{\left(1.05 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)} \\
&= 0.26 \times 10^7 \mathrm{~m}^{-1},\left[\because m_e C^2=0.51 \mathrm{MeV}\right] \\
& r_B\left(\text { Bohrs radius }=0.51 A=0.51 \times 10^{-10} \mathrm{~m}\right) \\
& \text { or } \lambda r_B=\left(0.26 \times 10^7 \mathrm{~m}^{-1}\right)\left(0.51 \times 10^{-10} \mathrm{~m}\right)=0.14 \times 10^{-13} \ll 1
\end{aligned}$

Further as $|F|=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left[\frac{1}{r^2+\frac{\lambda}{r}}\right] e^{-\lambda} r\ldots\ldots(i)$
and $|F|=\frac{d U}{d r}$

$\begin{aligned}
& U_r=\int|F| d r=\left(\frac{e^2}{4 \pi \epsilon_0}\right) \int\left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right) d r \\
& I f z=\frac{e^{-\lambda r}}{r}=\frac{1}{r}\left(e^{-\lambda r}\right) \\
& \frac{d z}{d r}=\left[\frac{1}{r}\left(e^{-\lambda r}\right)(-\lambda)+\left(e^{-\lambda r}\right)\left(\frac{1}{r^2}\right)\right]
\end{aligned}$

$\text { or } d z=-\left[\left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right)\right] d r$

Thus $\int\left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right) d r \Rightarrow-\int d z=-z=-\frac{e^{\lambda r}}{r}$ $=-\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{e^{\lambda r}}{r}\right) \ldots\ldots(ii)$

We know that,
$\begin{gathered}
m v \tau=h \Rightarrow v=\frac{h}{m m} ; \text { and } \\
\frac{m v^2}{r}=F=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{1}{r^2}+\frac{\lambda}{r}\right)
\end{gathered}$

$\left[\right.$ putting $e^{-\lambda r} \approx 1$ in eqn.(i)]
Thus $\left(\frac{m}{r}\right)\left(\frac{h^2}{4 \pi \epsilon_0}\right)=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{1}{r^2}+\frac{\lambda}{r}\right)$

or, $\frac{h^3}{m r^3}=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{r+\lambda r^2}{r^3}\right)$

or, $\frac{h^3}{m r^3}=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(r+\lambda r^2\right) \ldots\ldots(iii)$

when $\lambda=0, r=r_B^{\prime}$ and

$\frac{h^3}{m r^3}=\left(\frac{e^2}{4 \pi \epsilon_0}\right) r_B \ldots\ldots(iv)$

from eq.(iii) and (iv)

$\begin{aligned}
& r_B+r+\lambda r^2 \\
& \text { let } r=r+B+\delta \text { so that from }(i i i) \\
& r_B=\left(r_B+\delta\right)+\lambda\left(r_B^2+\delta^2+2 \delta r_B\right) \\
& \text { or } 0=\lambda r_B^2+\delta\left(1+2 \lambda r_B\right)\left(\text { neglectin } f \delta^2\right) \\
& \text { or } \delta=\frac{\lambda r_B^2}{\left(1+2 \lambda r_B\right)}=\left(-\lambda r_B^2\right)\left(1+\lambda r_B^2\right)^{-1} \\
& =\left(-\lambda r_B^2\right)\left(1+\lambda r_B^2\right)=\lambda r_B^2 \quad\left(\because \lambda r_B^2 \ll 1\right)
\end{aligned}$

from eg.(ii)

$\begin{aligned}
\mu_i&=-\left(\frac{e^2}{4 \pi \epsilon_0}\right) \frac{e^{-\lambda\left(r_B+\delta\right)}}{\left(r_B+\delta\right)} \\
& =-\left(\frac{e^2}{4 \pi \epsilon_0} \frac{1}{r_B}\right)\left(1-\frac{\delta}{r_B}\right)\left(1-\lambda r_B\right)\\
& =-\frac{e^2}{4 \pi \epsilon_0 r_B} \\
& =-24.2 e V
\end{aligned}$

$\left[\because e^{-\lambda\left(r_B+\bar{\sigma}\right)} \approx 1-\lambda\left(r_B-\delta\right)=1-\lambda r_B-\lambda \delta \approx 1-\lambda e_B\right]$

$\text { and } \frac{1}{\left(r_B+\delta\right)}=\frac{1}{r_B\left(1+\delta / r_B\right)}=\frac{1}{r_B}\left(1+\frac{\delta}{r_B}\right)^{-1}$

$=\frac{1}{r_B}\left(1-\frac{\delta}{r_B}\right)$

Further KE of the electron

$\begin{aligned}
K=&\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{h^2}{m^2 r^2}\right) \\
= & \frac{h^2}{2 m r^2}=\frac{h^2}{2 m\left(r_B+\delta\right)^2}=\frac{h^2}{2 m r_B^2+\left(1+\delta / r_B\right)^2} \\
= & \left(\frac{h^2}{2 m r_B^2}\right)\left(1+\frac{\delta}{r_B}\right)^{-2}=\left(\frac{h^2}{2 m r^2 B}\right)\left(1-\frac{2 \delta}{r_B}\right) \\
= & (13.6)\left(1+2 \lambda r_B\right) e V\left(a s \frac{h^2}{2 m r_B}=13.6 e V \text { and } \delta=-\lambda r_B^2\right)
\end{aligned}$

The total energy of H-atom in the ground state-final energy - initial energy

$\begin{aligned}
& =\left(-13.6+27.2 \lambda r_B\right) \mathrm{eV}-(-13.6 \mathrm{cV}) \\
& = \left(27.2 \lambda_{r_B}\right) \mathrm{eV}
\end{aligned}$

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 oppositecharge +q1, –q2 is modified to
$\left | F \right |=\frac{q_{1}q_{2}}{\left (4\pi \epsilon_{0} \right )}\frac{1}{r^{2}},r\geq R_{0}\\ =\frac{q_{1}q_{2}}{\left (4\pi \epsilon_{0} \right )}\frac{1}{R_{0}^{2}} \left ( \frac{R_{0}}{r} \right ),r\geq R_{0}\\$. Calculate in such a case, the ground state energy of an H-atom, if $\epsilon = 0.1, R_{0} = 1Å$

Answer:

Let us consider the case, when $\mathrm{r} \leq \mathrm{R}_0=1 A^o$
Let $\varepsilon=2+\delta$

$\mathrm{F}=\frac{q_1 q_2}{4 \pi \varepsilon_0} \cdot \frac{R_0^\delta}{r^{2+\bar{\delta}}}=x \frac{R_0^\delta}{r^{2+\kappa}}$

Where, $\frac{q_1 q_2}{4 \pi \varepsilon_0}=x=\left(1.6 \times 10^{19}\right)^2 \times 9 \times 10^9 =2.04 \times 10^{-29} N m^2$

The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons (Columbian force) provides the necessary centripetal force.

$\frac{m v^2}{r}=\frac{x R_0^\delta}{r^{2+\delta}} \text { or } v^2=\frac{x R_0^\delta}{m r^{1+\bar{\delta}}} \ldots\ldots(i)$

$m v r=n h \Rightarrow r=\frac{n h}{m v}=\frac{n h}{m}\left[\frac{m}{x R_0^\delta}\right]^{\frac{1}{2}} r^{\frac{1+\delta}{2}} \ldots$. [Applying Bhor's second postulates]
Solving this for r , we get $r_n=\left[\frac{n^2 h^2}{m x R_0^\delta}\right]^{\frac{1}{1-s}}$
Where $r_n$ is the radius of the $n^{\text {th }}$ orbit of the electron.
For $n=1$ and substituting the values of constant, we get

$\begin{aligned}
r_1&=\left[\frac{h^2}{m x R_0^6}\right]^{\frac{1}{1-s}} \\
\Rightarrow r_1&=\left[\frac{1.05^2 \times 10^{-68}}{9.1 \times 10^{-31} \times 2.3 \times 10^{-28} \times 10^{+} 19}\right]^{\frac{1}{35}} \\
& =8 \times 10^{-11} \\
& =0.08 \mathrm{~nm}(<0.1 \mathrm{~nm})
\end{aligned}$

This is the radius of the orbit of the electron in the ground state of the hydrogen atom. Again, using Bhor's second postulate, the speed of the electron

$v_n=\frac{n h}{m r_n}=n h\left(\frac{m x R_0^6}{n^2 h^2}\right)^{\frac{1}{1-\delta}}$

For $\mathrm{n}=1$, the speed of electron in ground state $v_1=\frac{h}{m r_1}=1.44 \times 10^6 \mathrm{~m} / \mathrm{s}$
The kinetic energy of electrons in the ground state

$\mathrm{K.E}=\frac{1}{2} m v_1^2-9.43 \times 10^{-19} J=5.9 \mathrm{eV}$

Potential energy of an electron in the ground state till $R_0$

$\mathrm{U}=\int_0^{R_0} F d r=\int_0^{R_0} \frac{x}{r^2} d r=-\frac{x}{R_0}$

Potential energy from $\mathrm{R}_0$ to $\mathrm{r}, \mathrm{U}=\int_{R_0}^r F d r=\int_{R_0}^r \frac{x R_0^6}{r^{2+6}} d r$

$\begin{aligned}
& \mathrm{U}=+x R_0^\delta \int_{R_0}^\delta \frac{d r}{r^{2+\delta}}=+\frac{x R_0^\delta}{-1-\delta}\left[\frac{1}{r^{1+\delta}}\right]_{R_0}^r \\
& \mathrm{U}=\frac{x R_0^\delta}{1+\delta}\left[\frac{1}{r^{1+\delta}}-\frac{1}{R_0^{1+\delta}}\right]=-\frac{x}{1+\delta}\left[\frac{R_0^\delta}{r^{1+\delta}}-\frac{1}{R_0}\right] \\
& \mathrm{U}=-x\left[\frac{R_0^\delta}{r^{1+\delta}}-\frac{1}{R_0}+\frac{1+\delta}{R_0}\right] \\
& \mathrm{U}=-x\left[\frac{R_0^{-19}}{r^{-0.9}}-\frac{1.9}{R_0}\right] \\
& =\frac{2.3}{0.9} \times 10^{-18}\left[(0.8)^{0.9}-1.9\right] J=-17.3 \mathrm{eV}
\end{aligned}$

Hence total energy of electron in ground state $=(-17.3+5.9)=-11.4 \mathrm{eV}$