Chapter 12 – Atoms Questions and Answers: NCERT Solutions for Class 12 Physics

1. Fill in the blanks using the given options:
1. The size of the atoms in Thomson’s model are _______ the atomic size in Rutherford’s model (much greater than/no different from/much lesser than).

Ans: The sizes of the atoms in Thomson’s model are no different from the atomic size in Rutherford’s model.

2. In the ground state of _______ electrons are in stable equilibrium, while in _______ electrons always experience a net force. (Thomson’s model/Rutherford’s model)

Ans: In the ground state of Thomson’s model, the electrons are in stable equilibrium, while in Rutherford’s model, electrons always experience a net force.

3. A classical atom based on _______ is doomed to collapse. (Thomson’s model/ Rutherford’s model.)

Ans: A classical atom based on Rutherford’s model is doomed to collapse.

4. An atom features a nearly continuous mass distribution in _______ but features a highly non- uniform mass distribution in _______ (Thomson’s model/ Rutherford’s model.)

Ans: An atom features a nearly continuous mass distribution in Thomson’s model, but features a highly non-uniform mass distribution in Rutherford’s model.

5. The positively charged part of the atom possesses most of the mass in _______ (Rutherford’s model/ Thomson’s model /both the models.)

Ans: The positively charged part of the atom possesses most of the mass in both the models.

2. If you’re given a chance to repeat the α−α−particle scattering experiment employing a thin sheet of solid hydrogen instead of the gold foil (Hydrogen is a solid at temperatures below 14 K). What results would you expect?

Ans: In the α−α−particle scattering experiment, when a thin sheet of solid hydrogen is replaced with the gold foil, the scattering angle would not turn out to be large enough.
This is because the mass of hydrogen is smaller than the mass of incident α−α−particles. Also, the mass of the scattering particle is more than the target nucleus (hydrogen).
As a consequence, the α−α−particles would not bounce back when solid hydrogen is utilized in the α−α−particle scattering experiment and hence, we cannot evaluate the size of the hydrogen nucleus.

3. What’s the shortest wavelength present within the Paschen series of spectral lines?

Ans: We know the Rydberg’s formula is given as;
hcλ=21.76×10−19[1n21−1n22]hcλ=21.76×10−19[1n12−1n22]
Here,
hh = Planck’s constant = 6.6×10−34Js6.6×10−34Js
cc = speed of light = 3×108m/s3×108m/s
(n1n1 and n2n2 are integers)
The shortest wavelength present within the Paschen series of the spectral lines is for values n1=3n1=3 and n2=∞n2=∞.
⇒⇒ hcλ=21.76×10−19[132−1∞2]hcλ=21.76×10−19[132−1∞2]
⇒⇒ λ=6.6×10−34×3×108×921.76×10−19λ=6.6×10−34×3×108×921.76×10−19
⇒⇒ λ=8.189×10−7mλ=8.189×10−7m
⇒⇒ λ=818.9nmλ=818.9nm

4. The two energy levels in an atom are separated by a difference of 2.3eV2.3eV. What is the frequency of radiation emitted when the atom makes a transition from the higher level to the lower level?

Ans: Given that the distance between the two energy levels in an atom is E=2.3eVE=2.3eV.
⇒E=2.3×1.6×10−19⇒E=2.3×1.6×10−19
⇒E=3.68×10−19J⇒E=3.68×10−19J
Let νν be the frequency of radiation emitted when the atom jumps from the upper level to the lower level.
The relation for energy is given as;
E=hνE=hν
Here,
h=h= Planck’s constant =6.6×10−34Js=6.6×10−34Js
⇒ν=Eh⇒ν=Eh
⇒ν=3.38×10−196.62×10−32⇒ν=3.38×10−196.62×10−32
⇒ν=5.55×1014Hz⇒ν=5.55×1014Hz
Clearly, the frequency of the radiation is 5.6×1014Hz5.6×1014Hz.

5. For a hydrogen atom, the ground state energy is −13.6eV−13.6eV . What are the kinetic and potential energies of the electron during this state?

Ans: Provided that the ground state energy of hydrogen atom, E=−13.6eVE=−13.6eVwhich is the total energy of a hydrogen atom.
Here, kinetic energy is equal to the negative of the total energy.
Kinetic energy =−E=−(−13.6)=13.6eV=−E=−(−13.6)=13.6eV
The potential energy is the same as the negative of two times kinetic energy.
Potential energy =−2×(13.6)=−27.2eV=−2×(13.6)=−27.2eV
∴∴ The kinetic energy of the electron is 13.6eV13.6eV and the potential energy is −27.2eV−27.2eV .

6. A hydrogen atom absorbs a photon when it is in the ground level, this excites it to the n=4n=4 level. Find out the wavelength and frequency of the photon.

Ans: It is known that for ground level absorption, n1=1n1=1
Let E1E1 be the energy of this level. It is known that E1E1 is related with n1n1 as;
E1=−13.6n21eVE1=−13.6n12eV
⇒E1=−13.612=−13.6eV⇒E1=−13.612=−13.6eV
When the atom jumps to a higher level, n2=4n2=4.
Let E2E2 be the energy of this level.
⇒E2=−13.6n22eV⇒E2=−13.6n22eV
⇒E2=−13.642=−13.616eV⇒E2=−13.642=−13.616eV
The amount of energy absorbed by the photon is given as;
E=E1−E2E=E1−E2
⇒E=(−13.616)−(−13.61)⇒E=(−13.616)−(−13.61)
⇒E=13.6×1516eV⇒E=13.6×1516eV
⇒E=13.6×1616×1.6×10−19⇒E=13.6×1616×1.6×10−19
⇒E=2.04×10−18J⇒E=2.04×10−18J
For a photon of wavelength λλ , the expression of energy is written as;
E=hcλE=hcλ
Here,
h=h= Planck’s constant =6.6×10−34Js=6.6×10−34Js
c=c= speed of light =3×108m/s=3×108m/s
⇒λ=hcE⇒λ=hcE
⇒λ=6.6×10−34×3×1082.04×10−18⇒λ=6.6×10−34×3×1082.04×10−18
⇒λ=9.7×10−8m⇒λ=9.7×10−8m
⇒λ=97nm⇒λ=97nm
Also, frequency of a photon is given by the relation,
ν=cλν=cλ
⇒ν=3×1089.7×10−8≈3.1×1015Hz⇒ν=3×1089.7×10−8≈3.1×1015Hz
Clearly, the wavelength of the photon is 97nm whereas the frequency is 3.1×1015Hz3.1×1015Hz.

7. Answer the following questions.
1. Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n=1,2n=1,2 and 33 levels.

Ans: Consider ν1ν1 to be the orbital speed of the electron in a hydrogen atom in the ground state level n1=1n1=1 . For charge (e)(e) of an electron, ν1ν1 is given by the relation,
ν1=e2n14π∈0(h2π)ν1=e2n14π∈0(h2π)
⇒ν1=e22∈0h⇒ν1=e22∈0h
Here,
e=1.6×10−19Ce=1.6×10−19C
∈0=∈0=Permittivity of free space =8.85×10−12N−1C2m−2=8.85×10−12N−1C2m−2
h=h= Planck’s constant =6.6×10−34Js=6.6×10−34Js
⇒ν1=(1.6×10−19)22×8.85×10−12×6.62×10−34⇒ν1=(1.6×10−19)22×8.85×10−12×6.62×10−34
⇒ν1=0.0218×108⇒ν1=0.0218×108
⇒ν1=2.18×106m/s⇒ν1=2.18×106m/s
For level n2=2n2=2 , we can write the relation for the corresponding orbital speed as;
ν2=e2n22∈0hν2=e2n22∈0h
⇒ν2=(1.16×10−19)22×2×8.85×10−12×6.62×10−34⇒ν2=(1.16×10−19)22×2×8.85×10−12×6.62×10−34
⇒ν2=1.09×106m/s⇒ν2=1.09×106m/s
And, for n3=3n3=3 , we can write the relation for the corresponding orbital speed as;
ν3=e2n32∈0hν3=e2n32∈0h
⇒ν3=(1.16×10−19)23×2×8.85×10−12×6.62×10−34⇒ν3=(1.16×10−19)23×2×8.85×10−12×6.62×10−34
⇒ν3=7.27×105m/s⇒ν3=7.27×105m/s
Clearly, the speeds of the electron in a hydrogen atom in the levels n=1,2n=1,2 and 33 are 2.18×106m/s2.18×106m/s , 1.09×106m/s1.09×106m/s and 7.27×105m/s7.27×105m/s respectively.

2. Calculate the orbital period in each of these levels.

Ans: Consider T1T1 to be the orbital period of the electron when it is in level n1=1n1=1 .
It is known that the orbital period is related to the orbital speed as
T1=2πr1ν1T1=2πr1ν1
Here,
r1=r1= Radius of the orbit in n1n1=n21h2∈0πme2=n12h2∈0πme2
h=h= Planck’s constant =6.6×10−34Js=6.6×10−34Js
e=e= Charge of an electron =1.6×10−19C=1.6×10−19C
∈0=∈0= Permittivity of free space =8.85×10−12N−1C2m−2=8.85×10−12N−1C2m−2
m=m= Mass of an electron =9.1×10−31kg=9.1×10−31kg
⇒T1=2π×(1)2×(6.62×10−34)2×8.85×10−122.18×106×π×9.1×10−31×(1.6×10−19)2⇒T1=2π×(1)2×(6.62×10−34)2×8.85×10−122.18×106×π×9.1×10−31×(1.6×10−19)2
⇒T1=15.27×10−17⇒T1=15.27×10−17
⇒T1=1.527×10−16s⇒T1=1.527×10−16s
For level n2=2n2=2, we can write the orbital period as;
T2=2πr2ν2T2=2πr2ν2
Here,
r2=r2= Radius of the orbit in n2n2=n22h2∈0πme2=n22h2∈0πme2
⇒T1=2π×(2)2×(6.62×10−34)2×8.85×10−121.09×106×π×9.1×10−31×(1.6×10−19)2=1.22×10−15s⇒T1=2π×(2)2×(6.62×10−34)2×8.85×10−121.09×106×π×9.1×10−31×(1.6×10−19)2=1.22×10−15s
And for the level n3=3n3=3, we can write the orbital period as;
T3=2πr3ν3T3=2πr3ν3
Here,
r3=r3= Radius of the orbit in n3n3=n23h2∈0πme2=n32h2∈0πme2
⇒T3=2π×(3)2×(6.62×10−34)2×8.85×10−127.27×105×π×9.1×10−31×(1.6×10−19)2=4.12×10−15s⇒T3=2π×(3)2×(6.62×10−34)2×8.85×10−127.27×105×π×9.1×10−31×(1.6×10−19)2=4.12×10−15s
Hence, the orbital periods in the levels n=1,2n=1,2 and 33 are 1.527×10−16s1.527×10−16s , 1.22×10−15s1.22×10−15s and 4.12×10−15s4.12×10−15s respectively.

8. The innermost electron orbit of a hydrogen atom has a radius of 5.3×10−11m5.3×10−11m. What are the radii of the n=2n=2 and n=3n=3 orbits?

Ans: Provided that the innermost radius, r1=5.3×10−11mr1=5.3×10−11m .
Let r2r2 be the radius of the orbit at n=2n=2 . It is related to the radius of the innermost orbit as;
r2=(n)2r1r2=(n)2r1
⇒r2=(2)2×5.3×10−11=2.1×10−10m⇒r2=(2)2×5.3×10−11=2.1×10−10m
Similarly, for n=3n=3;
r3=(n)2r1r3=(n)2r1
⇒r3=(3)2×5.3×10−11=4.77×10−10m⇒r3=(3)2×5.3×10−11=4.77×10−10m
Clearly, the radii of the n=2n=2 and n=3n=3 orbits are 2.1×10−10m2.1×10−10m and 4.77×10−10m4.77×10−10m respectively.

9. A 12.5eV12.5eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Ans: It is provided that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5eV12.5eV .
It is also known that the energy of the gaseous hydrogen in its ground state at room temperature is −13.6eV−13.6eV .
When gaseous hydrogen is bombarded with an electron beam at room temperature, the energy of the gaseous hydrogen becomes
−13.6+12.5eV=−1.1eV−13.6+12.5eV=−1.1eV
.
Now, the orbital energy is related to orbit level
(n)(n)
as;
E=−13.6(n)2eVE=−13.6(n)2eV
For n=3;E=−13.69=−1.5eVn=3;E=−13.69=−1.5eV
This energy is approximately equal to the energy of gaseous hydrogen.
So, it can be concluded that the electron has excited from
n=1ton=3n=1ton=3
level.
During its de-excitation, the electrons can jump from
n=3ton=1n=3ton=1
directly, which forms a line of the Lyman series of the hydrogen spectrum.
The formula for wave number for Lyman series is given as;
1λ=Ry(112−1n2)1λ=Ry(112−1n2)
Here,
Ry=Ry= Rydberg constant =1.097×107m−1=1.097×107m−1
λ=λ=
Wavelength of radiation emitted by the transition of the electron
Using this relation for n=3n=3 we get,
1λ=1.097×107(112−132)1λ=1.097×107(112−132)
⇒1λ=1.097×107(1−19)⇒1λ=1.097×107(1−19)
⇒1λ=1.097×107(89)⇒1λ=1.097×107(89)
⇒λ=98×1.097×107=102.55nm⇒λ=98×1.097×107=102.55nm
If the transition takes place from n=3ton=2n=3ton=2 , and then from
n=2ton=1n=2ton=1
, then the wavelength of the radiation emitted in transition from
n=3ton=2n=3ton=2
is given as;
1λ=1.097×107(122−132)1λ=1.097×107(122−132)
⇒1λ=1.097×107(14−19)⇒1λ=1.097×107(14−19)
⇒1λ=1.097×107(536)⇒1λ=1.097×107(536)
⇒λ=365×1.097×107=656.33nm⇒λ=365×1.097×107=656.33nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Now, the wavelength of the radiation when the transition takes place from
n=2ton=1n=2ton=1
is given as;
1λ=1.097×107(112−122)1λ=1.097×107(112−122)
⇒1λ=1.097×107(1−14)⇒1λ=1.097×107(1−14)
⇒1λ=1.097×107(34)⇒1λ=1.097×107(34)
⇒λ=43×1.097×107=121.54nm⇒λ=43×1.097×107=121.54nm
Clearly, in the Lyman series, two wavelengths are emitted i.e., 102.5nm and 121.5nm whereas in the Balmer series, only one wavelength is emitted i.e., 656.33nm.

10. In accordance with the Bohr’s model, what is the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5×1011m1.5×1011m with an orbital speed of 3×104m/s3×104m/s . The mass of the earth is given as 6×1024kg6×1024kg .

Ans: Here, it is provided that,
Radius of the earth’s orbit around the sun, r=1.5×1011mr=1.5×1011m
Orbital speed of the earth, ν=3×104m/sν=3×104m/s
Mass of the earth, m=6×1024kgm=6×1024kg
With respect to the Bohr’s model, angular momentum is quantized and is given as;
mνr=nh2πmνr=nh2π
Here,
h=h= Planck’s constant =6.6×10−34Js=6.6×10−34Js
n=n= Quantum number
n=mνr2πhn=mνr2πh
⇒n=2π×6×1024×3×104×1.5×10116.62×10−34⇒n=2π×6×1024×3×104×1.5×10116.62×10−34
⇒n=25.61×1073=2.6×1074⇒n=25.61×1073=2.6×1074
Clearly, the quantum number that characterizes the earth’s revolution around the sun is 2.6×10742.6×1074 .

11. Which of the following questions help you understand the difference between Thomson’s model and Rutherford’s model better.
1. Is the average angle of deflection of α−α−particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

Ans: About the same.
The average angle of deflection of α−α−particles caused by a thin gold foil considered by Thomson’s model is about the same size as that considered by Rutherford’s model.
This is because in both the models, the average angle was used.

2. Is the probability of backward scattering (i.e., scattering of α−α−particles at angles greater than 90∘90∘) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

Ans: Much less.
The probability of scattering of α−α−particles at angles greater than 90° considered by Thomson’s model is much less than that considered by Rutherford’s model.

3. Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α−α−particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

Ans: Scattering is mainly caused due to single collisions.
The chances of a single collision have a linearly increasing nature with the number of target atoms.
As the number of target atoms increases with an increase in thickness, the collision probability varies linearly with the thickness of the target.

4. In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α−α−particles by a thin foil?

Ans: Thomson’s model.
It is incorrect to not consider multiple scattering in Thomson’s model for the calculation of average angle of scattering of α−α−particles by a thin foil.
This is because a single collision produces very little deflection in this model.
Thus, the observed average scattering angle can be demonstrated only by considering multiple scattering.

12. The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−4010−40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Ans: The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−4010−40.
It is given that,
Radius of the first Bohr orbit is given by the relation,
r1=4π∈0(h2π)2mee2r1=4π∈0(h2π)2mee2 …(1)
Here,
∈0=∈0= Permittivity of free space
h=h= Planck’s constant =6.63×10−34Js=6.63×10−34Js
me=me= Mass of an electron =9.1×10−31kg=9.1×10−31kg
e=e= Charge of an electron =1.6×10−19C=1.6×10−19C
mp=mp= Mass of a proton =1.67×10−27kg=1.67×10−27kg
r=r= Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given by;
FC=e24π∈0r2FC=e24π∈0r2 …(2)
Gravitational force of attraction between an electron and a proton is given by;
FG=Gmpmer2FG=Gmpmer2 …(3)
Here,
G=G= Gravitational constant =6.67×10−11Nm2/kg2=6.67×10−11Nm2/kg2
The electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write;
FG=FCFG=FC
⇒Gmpmer2=e24π∈0r2⇒Gmpmer2=e24π∈0r2 …(using 2 & 3)
⇒e24π∈0=Gmpme⇒e24π∈0=Gmpme …(4)
Now, by substituting the value of equation (4) in (1), we get;
r1=(h2π)2Gmpmer1=(h2π)2Gmpme
⇒r1=(6.63×10−342×3.14)26.67×10−11×1.67×10−27×(9.1×10−31)2⇒r1=(6.63×10−342×3.14)26.67×10−11×1.67×10−27×(9.1×10−31)2
⇒r1≈1.21×1029m⇒r1≈1.21×1029m
It is known that the universe is 156 billion light years wide or 1.5×1027m1.5×1027m wide. Clearly, it can be concluded that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

13. Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level nn to level (n−1)(n−1). For large nn , show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Ans: It is given that a hydrogen atom makes transition from an upper level (n)(n) to a lower level (n−1)(n−1). We have the relation for energy (E1)(E1) of radiation at level nn as;
E1=hν1=hme4(4π)3∈20(h2π)3×(1n2)E1=hν1=hme4(4π)3∈02(h2π)3×(1n2) …(1)
Here,
ν1=ν1= Frequency of radiation at level nn
h=h= Planck’s constant
m=m= Mass of hydrogen atom
e=e= Charge on an electron
∈0=∈0= Permittivity of free space
Now, the relation for energy (E2)(E2) of radiation at level (n−1)(n−1) is given as;
E2=hν2=hme4(4π)3∈20(h2π)3×1(n−1)2E2=hν2=hme4(4π)3∈02(h2π)3×1(n−1)2 …(2)
Here,
ν2=ν2= Frequency of radiation at level (n−1)(n−1)
Energy (E)(E) released as a result of de-excitation;
E=E2−E1E=E2−E1
⇒hν=E2−E1⇒hν=E2−E1 …(3)
Here,
ν=ν= Frequency of radiation emitted
Using the values of equations (1) and (2) in equation (3), we get;
ν=me4(4π)3∈20(h2π)3[1(n−1)2−1n2]ν=me4(4π)3∈02(h2π)3[1(n−1)2−1n2]
⇒ν=me4(2n−1)(4π)3∈20(h2π)3n2(n−1)2⇒ν=me4(2n−1)(4π)3∈02(h2π)3n2(n−1)2
For large nn, we can write (2n−1)−2n(2n−1)−2n and (n−1)≃n(n−1)≃n
⇒ν=me432π3∈20(h2π)3n3⇒ν=me432π3∈02(h2π)3n3 …(4)
Classical relation of frequency of revolution of an electron is given by;
νc=ν2πrνc=ν2πr …(5)
Here,
Velocity of the electron in the nthnth orbit is given as;
ν=e24π∈0(h2π)nν=e24π∈0(h2π)n …(6)
And, radius of the nthnth orbit is given by;
r=4π∈0(h2π)2me2n2r=4π∈0(h2π)2me2n2 …(7)
Substituting the values of equation (6) and (7) in equation (5), we get;
νc=me432π3∈20(h2π)3n3νc=me432π3∈02(h2π)3n3 …(8)
Clearly, when equations (4) and (8) are compared, it can be seen that the frequency of radiation emitted by the hydrogen atom is the same as the classical orbital frequency.

14. Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (∼10−10m)(∼10−10m).
1. Construct a quantity with the dimensions of length from the fundamental constants e,meandce,meandc. Also, determine its numerical value.

Ans: To construct a quantity with the dimensions of length from the fundamental constants e,meandce,meandc, take:
Charge of an electron, e=1.6×10−19Ce=1.6×10−19C
Mass of an electron, me=9.1×10−31kgme=9.1×10−31kg
Speed of light, c=3×108m/sc=3×108m/s
The quantity having dimensions of length and involving the given quantities is (e24π∈0mec2)(e24π∈0mec2)
Here,

∈0=∈0= Permittivity of free space
And, 14π∈0=9×109Nm2C−214π∈0=9×109Nm2C−2
∴∴The numerical value of the taken quantity will be
14π∈0×e2mec2=9×109×(1.6×10−19)29.1×10−31×(3×108)2=2.81×10−15m14π∈0×e2mec2=9×109×(1.6×10−19)29.1×10−31×(3×108)2=2.81×10−15m
Clearly, the numerical value of the taken quantity is much less than the typical size of an atom.

2. You will observe that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for “something else” to get the right atomic size. Now, the Planck’s constant hh had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h,⊸me,⊸and⊸eh,⊸me,⊸and⊸e
will yield the right atomic size. Construct a quantity with the dimension of length from h,meandeh,meande
and confirm that its numerical value has indeed the correct order of magnitude.

Ans: To construct a quantity with the dimension of length from
h,meandeh,meande
, take,
Charge of an electron, e=1.6×10−19Ce=1.6×10−19C
Mass of an electron, me=9.1×10−31kgme=9.1×10−31kg
Planck’s constant, h=6.63×10−34Jsh=6.63×10−34Js
Now, let us take a quantity involving the given quantities as, 4π∈0(h22π)mee24π∈0(h22π)mee2
Here, ∈0=∈0= Permittivity of free space
And, 14π∈0=9×109Nm2C−214π∈0=9×109Nm2C−2
∴∴The numerical value of the taken quantity would be
4π∈0×(h2π)2mee2=19×109×(6.63×10−342×3.14)29.1×10−31×(1.6×10−19)2=0.53×10−10m4π∈0×(h2π)2mee2=19×109×(6.63×10−342×3.14)29.1×10−31×(1.6×10−19)2=0.53×10−10m
Clearly, the value of the quantity taken is of the order of the atomic size.

15. The total energy of an electron in the first excited state of the hydrogen atom is about −3.4eV−3.4eV.
1. What is the kinetic energy of the electron in this state?

Ans: Kinetic energy of the electron is the same as the negative of the total energy.
⇒K=−E⇒K=−E
⇒K=−(−3.4)=+3.4eV⇒K=−(−3.4)=+3.4eV
Clearly, the kinetic energy of the electron in the given state is +3.4eV+3.4eV.

2. What is the potential energy of the electron in this state?

Ans: Potential energy (U)(U) of the electron is the same as the negative of twice of its kinetic energy
⇒U=−2K⇒U=−2K
⇒U=−(−3.4)=−6.8eV⇒U=−(−3.4)=−6.8eV
Clearly, the potential energy of the electron in the given state is −6.8eV−6.8eV.

3. Which of the answers above would change if the choice of the zero of potential energy is changed?

Ans: The potential energy of a system is dependent on the reference point taken. Here, the potential energy of the reference point is considered to be zero.
When the reference point is changed, then the magnitude of the potential energy of the system also changes.
As total energy is the sum of kinetic and potential energies, total energy of the system would also differ.

16. If Bohr’s quantization postulate (angular momentum =nh/2n=nh/2n) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun?

Ans: It is not much spoken about the quantization of orbits of planets around the Sun since the angular momentum associated with planetary motion is largely relative to the value of constant (h)(h).
The angular momentum of the Earth in its orbit is of the order of 1070h1070h. This causes a very high value of quantum levels nn of the order of 10701070.
When large values of nn are considered, successive energies and angular momenta are found to be relatively very small. Clearly, the quantum levels for planetary motion are considered continuous.

17. Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom (i.e., an atom in which a negatively charged muon (μ−)(μ−)mass about 207me207me. orbits around a proton).

Ans: To obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom, consider the mass of a negat ively charged muon to be mμ=207mmμ=207m.
Now, according to Bohr’s model,
Bohr radius, re∝(1me)re∝(1me)
And, energy of a ground state electronic hydrogen atom, Ee∝meEe∝me.
Also, energy of a ground state muonic hydrogen atom, Eμ∝meEμ∝me.
It is known that the value of the first Bohr orbit, re=0.53A=0.53×10−10mre=0.53A=0.53×10−10m
Consider rμrμ to be the radius of muonic hydrogen atom.
At equilibrium, we have the relation:
mμrμ=meremμrμ=mere
⇒207me×rμ=mere⇒207me×rμ=mere
⇒rμ=0.53×10−10207=2.56×10−13m⇒rμ=0.53×10−10207=2.56×10−13m
Clearly, the value of the first Bohr radius of muonic hydrogen atom is 2.56×10−13m2.56×10−13m.
Now, we have,
Ee=−13.6eVEe=−13.6eV
Taking the ratio of these energies as EeEμ=memμEeEμ=memμ
⇒EeEμ=me207me⇒EeEμ=me207me
⇒Eμ=207Ee⇒Eμ=207Ee
⇒Eμ=207×(−13.6)=−2.81keV⇒Eμ=207×(−13.6)=−2.81keV
Clearly, the ground state energy of a muonic hydrogen atom is −2.81keV−2.81keV.