NCERT Solutions for Class 12 Physics Chapter 12 Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms is provided here in a comprehensive way. In class 12 NCERT physics book, chapter 12 is about atoms where students learn about advanced concepts of atoms such as Rutherford’s model, Paschen series of spectral lines, etc. Questions related to this chapter is common in CBSE board exam and in competitive exams like NEET and JEE.

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Studying these solutions will make you perfect in answering textbook questions along with few exemplary questions, MCQ’S, numerical problems, worksheets and atoms class 12 important questions. It is important for the students to class 12 physics chapter atoms notes in order to save time while they revise for the examination.

Class 12 Physics NCERT Solutions for Chapter 12 Atom

NCERT Class 12 physics solutions for Chapter 12 Atoms given here are very simple and easy to understand. These solutions can help the students to easily clear their doubts and have a proper understanding of the fundamentals.

Topics covered in Class 12 Physics Chapter 12 Atom are:

Section Number Topic
12.1 Introduction
12.2 Alpha-Particle Scattering And Rutherford’s Nuclear Model Of Atom
12.2.1 Alpha-particle trajectory
12.2.2 Electron orbits
12.3 Atomic Spectra
12.3.1 Spectral series
12.4 Bohr Model Of The Hydrogen Atom
12.4.1 Energy levels
12.5 The Line Spectra Of The Hydrogen Atom
12.6 De Broglie’s Explanation Of Bohr’s Second Postulate Of Quantisation.

Define Rutherford Atomic Model?

Everything in the universe is composed of atoms. An atom is the fundamental building block of all matter. The Rutherford atomic model was proposed by Ernest Rutherford. In this model, the atom is described as a minute, dense, positively charged core called nucleus, around which negatively charged constituents called electrons revolve, much like the planets revolving around the sun.

 

Class 12 physics NCERT Solutions Chapter 12 Atom Important Questions


Q1: Choose the correct alternative from the clues given at the end of each statement:

(a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)

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

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

(d) An atom has a nearly continuous mass distribution in a ……….but has a highly non-uniform mass distribution in ……….(Thomson’s model/ Rutherford’s model.)

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

Ans:

(a)  The side of an atom in Rutherford’s model and of Thomson’s model is the same.

(b) During stable equilibrium, electrons of Rutherford’s model experience net force while the electrons of Thomson’s model do not.

(c) Rutherford’s model is a showcases a model of an atom which is bound to fail.

(d) Mass distribution is uniform in Thomson’s model but highly irregular in Rutherford’s model.

(e) Both of the models state that a larger portion of the mass of an atom is contributed by the positively charged part.

Q2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Ans:

We know that the mass of the incident alpha particle (6.64 × 10-27kg) is more than the mass of hydrogen (1.67 × 1027Kg). Hence, the target nucleus is lighter, from which we can conclude that the alpha particle would not rebound. Implying to the fact that solid hydrogen isn’t a suitable replacement to gold foil for the alpha particle scattering experiment.

Q3: What is the shortest wavelength present in the Paschen series of spectral lines?

Ans:

By Rydberg’s formula :

hcλ=21.76×1019[1n121n22]\frac{hc}{\lambda } = 21.76 \times 10^{-19}\left [ \frac{1}{{n_{1}}^{2}}-\frac{1}{{n_{2}}^{2}} \right ]\\

Planck’s constant h = 6.6 × 10-34 Js

Speed of light c = 3 × 108 m/s

n1 and n2 are integers.

To obtain shortest wavelength we substitute n1 = 3 and n2= \infty hcλ=21.76×1019[1(3)21()2]\frac{hc}{\lambda } = 21.76 \times 10^{-19}\left [ \frac{1}{{(3)}^{2}}-\frac{1}{{(\infty )}^{2}} \right ]\\

i.e.   λ=6.6×1034×3×108×921.76×1019\\\lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 9}{21.76 \times 10^{-19}}\\ = 8.189 × 10 -7 = 818.9 nm

Q4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Ans:

Separation of two energy levels in an atom, E = 4.6 eV = 4.6 × 1.6 × 10−19 = 7.36 × 10−19 J

Consider ν as the frequency of radiation emitted when the atom transitions from the upper level

to the lower level.

Following is the relation for energy:

E = hv

Where, h = Planck’s constant = 6.62 × 10-34 Js

v=Eh=7.36×10196.62×1032\mathit{v} = \frac{E}{h} = \frac{7.36 \times 10^{-19}}{6.62 \times 10^{-32}} = 11.1 × 1014 Hz

Hence, the frequency of the radiation is 11.1 × 1014 Hz.

Q5: The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Ans:

Ground state energy of hydrogen atom, E = − 27.2 eV

The total energy of hydrogen atom is -27.2eV. The kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 27.2) = 27.2 eV

Potential energy = negative of two times of kinetic energy.

Potential energy = − 2 × ( 27.2 ) = − 54.4 eV

Q6: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Ans:

For ground level, n1= 1

Let, E1 be the energy of this level. It is known that E2 is related with n1 as:

E1=13.6(n1)2E_{1}=\frac{-13.6}{{(n_{1})^{2}}} eV =13.612=\frac{-13.6}{1^{2}} = 13.6 eV

The atom is excited to a higher level, n2= 3

Let, E2 be the energy of this level:

E2=13.6(n2)2E_{2}=\frac{-13.6}{{(n_{2})^{2}}} eV =13.632=\frac{-13.6}{3^{2}} =13.69=\frac{-13.6}{9} eV

Following is the amount of energy absorbed by the photon:

E = E2− E1

=13.69(13.6)=\frac{-13.6}{9}-(-13.6) =13.6×89=\frac{13.6 \times 8}{9} eV =13.6×89×1.6×1019=\frac{13.6 \times 8}{9} \times 1.6 \times 10^{-19} = 1.934 x 10-18 J

For a photon of wavelength λ, the expression of energy is written as:

E=hcλE = \frac{hc}{\lambda }

Where,

h = Planck’s constant = 6.6 × 10−34 Js

c = Speed of light = 3 × 108 m/s

λ=hcE\lambda = \frac{hc}{E} =6.6  ×  1034  ×  3  ×  1081.934  ×  1018= \frac{6.6 \;\times \;10^{-34} \;\times \;3 \;\times \;10^{8}}{1.934 \;\times \;10^{-18}}\\ = 10.23 x 10-8 m = 102.3 nm

And, frequency of a photon is given by the relation:

v=cλ\mathit{v}=\frac{c}{\lambda } =3×10810.23×108= \frac{3\times 10^8}{10.23\times 10^{-8}} = 2.93 x 1015 Hz

Therefore, 102.3nm and 2.93×1015Hz is the wavelength and frequency of the photon.

Q.7: (a) (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Ans:

(a) Let v1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1= 1. For charge (e) of an electron, v1 is given by the relation,

v1=e2n14  π  ϵ0  (h2π)=e22  ϵ0  hv_1 = \frac{e^{2}}{n_1 4\;\pi\;\epsilon_0\; (\frac{h}{2\pi}) } = \frac{e^{2}}{2\;\epsilon_0 \;h}\\

Where, e = 1.6 × 10−19 C

ϵ0\epsilon_0 = Permittivity of free space = 8.85 × 10−12 N−1 C 2 m−2

h = Planck’s constant = 6.62 × 10−34 Js

v1=(1.6×1019)22×8.85×1012×6.62×1034v_1 = \frac{(1.6\times10^{-19})^{2}}{2\times8.85\times10^{-12}\times6.62\times10^{-34}} = 0.0218 x 108 = 2.18 × 106 = 106 m/s

For level n2= 2, we can write the relation for the corresponding orbital speed as:

v2=e2n22ϵ0hv_2 = \frac{e^2}{n_2 2 \epsilon_0 h} =(1.6×1019)22×2×8.85×1032×6.62×1034= \frac{(1.6 \times 10^{-19})^2}{2 \times 2 \times 8.85 \times 10^{-32} \times 6.62 \times 10^{-34}} = 1.09 × 106 m/s

Therefore, in a hydrogen atom, the speed of the electron at different levels that is n=1, n=2, and n=3 is 2.18×106m/s and 1.09×106m/s.

(b) Let, T1 be the orbital period of the electron when it is in level n1= 1.

Orbital period is related to orbital speed as:

T1=2πr1v1T_1 = \frac{ 2 \pi r_1 } { v_1 }   [Where, r1 = Radius of the orbit] =n12  h2  ϵ0π  me2=\frac{ { n _ 1 } ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\\

h = Planck’s constant = 6.62 × 10−34 Js

e = Charge on an electron = 1.6 × 10−19 C

0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2

m = Mass of an electron = 9.1 × 10−31 kg

T1=2πr1v1T_1 = \frac{ 2 \pi r_1 } { v_1 } =2  π  ×(2)2  ×  (6.62  ×  1034)2  ×  8.85  ×  10122.18  ×  106  ×  π  ×9.1  ×  1031  ×  (1.6  ×  1019)2=\frac{2 \;\pi \;\times ( 2 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 } \;\times \;8.85 \;\times \;10 ^{ -12 }}{ 2.18 \;\times \;10 ^ { 6 } \;\times \;\pi \;\times 9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } } = 15.27 × 10-17 = 1.527 × 10-16 s

For level n2 = 2, we can write the period as:

T2=2πr2v2T_2 = \frac{ 2 \pi r_2 } { v_2 }   [Where, r2 = Radius of the electron in n2 = 2] =(n2)2  h2  ϵ0π  me2=\frac{ {( n _ 2 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }  =2  π  ×  (2)2  ×  (6.62  ×  1034)2  ×  8.85  ×10121.09  ×  106  ×  π  ×  9.1  ×  1031  ×  (1.6  ×  1019)2=\frac{2 \;\pi \;\times \;( 2 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 }\;\times \;8.85 \;\times 10 ^{ -12 }}{ 1.09 \;\times \;10 ^ { 6 } \;\times \;\pi \;\times \;9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } } = 1.22 × 10-15s

Therefore, 1.52×10-16s and 1.22×10-15s are the orbital periods in each levels.

Q8: The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?

Ans:

The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 × 10−11 m.

Let r2 be the radius of the orbit at n = 2. It is related to the radius of the inner most orbit as:

r2=(n)2r1r_2 = ( n ) ^ { 2 } r_ 1 = 4 × 5.3 × 10-11 = 2.12 × 10-10

For n = 3, we can write the corresponding electron radius as:

r3=(n)2r1r_3 = ( n ) ^ { 2 } r_ 1 = 9 × 5.3 x 10-11 = 4.77 × 10-10

Therefore, 2.12 × 10−10 m and 4.77 × 10−10 m are the radii of an electron for n = 2 and n = 3 orbits are respectively.

Q9: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Ans:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

E=13.6n2  eVE = \frac {-13.6 } { n ^ { 2 } } \; eV\\

For n = 3, E = -13.6 / 9 = – 1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen.

It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly,

which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

1λ=Ry(1121n2)\frac { 1 } { \lambda } = R_y \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { n ^ { 2 } } \right )

Where, Ry= Rydberg constant = 1.097 × 107 m−1

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

1λ=1.097×107(112132)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right ) 1λ=1.097×107(119)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 9 } \right ) 1λ=1.097×107(89)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 8 } { 9 } \right )\\ λ=98×1.097×107\\\lambda = \frac { 9 }{8 \times 1.097 \times 10 ^ {7 }} = 102.55 nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given

as:

1λ=1.097×107(112122)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 2 ^ { 2 } } \right ) 1λ=1.097×107(114)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 4 } \right ) 1λ=1.097×107(34)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 3 } { 4 } \right )\\ λ=43×1.097×107\\\lambda = \frac { 4 }{3 \times 1.097 \times 10 ^ {7 }} = 121.54 nm

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is

given as:

1λ=1.097×107(122132)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 2 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right ) 1λ=1.097×107(1419)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 4 } -\frac { 1 } { 9 } \right ) 1λ=1.097×107(536)\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 5 } { 36 } \right )\\ λ=365×1.097×107\\\lambda = \frac { 36 }{ 5 \times 1.097 \times 10 ^ {7 }} = 656.33 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Therefore, there are two wavelengths that are emitted in Lyman series and they are 102.5 nm and 121.5 nm and one wavelength in the Balmer series which is 656.33 nm.

Q10: In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of diameter 3 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)

Ans:

Radius of the orbit of the Earth around the Sun, r = 1.5 × 1011 m

Orbital speed of the Earth, ν = 3 × 104 m/s

Mass of the Earth, m = 6.0 × 1024 kg

According to Bohr’s model, angular momentum is quantumized and given as:

Where, mvr=nh2πm v r = \frac { n h } { 2 \pi }

h = Planck’s constant = 6.62 × 10−34 Js

n = Quantum number

n=mvr2πhn = \frac { m v r 2 \pi } { h } =2π×6×1024×3×104×1.5×10116.62×1034= \frac{ 2 \pi \times 6 \times 10^{24} \times 3 \times 10 ^ {4} \times 1.5 \times 10^{11} }{ 6.62 \times 10^{-34} } = 25.61 ×1073 = 2.6 × 1074

Therefore, the earth revolution can be characterized by the quantum number 2.6 × 1074.

 Additional Questions:

Q11: Choose a suitable solution to the given statements which justify the difference between Thomson’s model and Rutherford’s model

(a) In the case of scattering of alpha particles by a gold foil, average angle of deflection of alpha particles stated by Rutherford’s model is( less than, almost the same as, much greater than )stated by Thomson’s model.

(b) Is the likelihood of reverse scattering (i.e., dispersing of α-particles at points more prominent than 90°) anticipated by Thomson’s model ( considerably less, about the same, or much  more prominent ) than that anticipated by Rutherford’s model?

(c) For a small thickness T, keeping other factors constant, it has been found that amount of alpha particles scattered at direct angles is proportional to T. This linear dependence implies?

(d) To calculate average angle of scattering of alpha particles by thin gold foil, which model states its wrong to skip multiple scattering?

Ans:

(a) almost the same

The normal point of diversion of alpha particles by a thin gold film anticipated by Thomson’s model is about the same as from anticipated by Rutherford’s model. This is on the grounds that the average angle was taken in both models.

(b) much less

The likelihood of scattering of alpha particles at points more than 90° anticipated by Thomson’s model is considerably less than that anticipated by Rutherford’s model.

(c) Dispersing is predominantly because of single collisions. The odds of a single collision increment linearly with the amount of target molecules. Since the number of target particles increment with an expansion in thickness, the impact likelihood depends straightly on the thickness of the objective.

(d) Thomson’s model

It isn’t right to disregard multiple scattering in Thomson’s model for figuring out the average angle of scattering of alpha particles by a thin gold film. This is on the grounds that a solitary collision causes almost no deflection in this model. Subsequently, the watched normal scattering edge can be clarified just by considering multiple scattering.

Q12: The gravitational attraction amongst proton and electron in a hydrogen atom is weaker than the coulomb attraction by a component of around 10−40. Another option method for taking a gander at this case is to assess the span of the first Bohr circle of a hydrogen particle if the electron and proton were bound by gravitational attraction. You will discover the appropriate response fascinating.

Ans:

Radius of the first Bohr orbit is given by the relation:

r1=4πϵ0(h2π)2mee2r_1 = \frac{4 \pi \epsilon_0 \left ( \frac{h}{2 \pi} \right )^2}{m_e e^2 }       —–(1)

Where,

0 = Permittivity of free space

h = Planck’s constant = 6.63 × 10−34 Js

me = Mass of an electron = 9.1 × 10−31 kg

e = Charge of an electron = 1.9 × 10−19 C

mp = Mass of a proton = 1.67 × 10−27 kg

r = Distance between the electron and the proton

Coulomb attraction between an electron and a proton is:

FC=e24πϵ0r2F_C = \frac{e^{2}}{4 \pi \epsilon_0 r^{2}}—–(2)

Gravitational force of attraction between an electron and a proton is:

FG=Gmempr2F_G = \frac{G m_e m_p}{r^{2}}——(3)

Where,

G = Gravitational constant = 6.67 × 10−11 N m2/kg2

Considering electrostatic force and the gravitational force between an electron and a proton to be equal, we get:

∴ FG = FC

Gmempr2=e24πϵ0r2\frac{G m_e m_p}{r^{2}}=\frac{e^{2}}{4 \pi \epsilon_0 r^{2}}\\ Gmemp=e24πϵ0{G m_e m_p}=\frac{e^{2}}{4 \pi \epsilon_0 }—–(4)

Putting the value of equation (4) in equation (1), we get:

r1=(h2π)2Gmempr1 = \frac{\left ( \frac{h}{2 \pi} \right )^2}{G m_e m_p} r1=(6.63×10342×3.14)26.67×1011×1.67×1027×(9.1×1031)2r1 = \frac{\left ( \frac{6.63\times10^{-34}}{2 \times 3.14} \right )^2}{6.67\times 10^{-11}\times 1.67 \times 10^{-27}\times(9.1\times 10^{-31})^{2} } = 1.21 × 1029 m

It is known that the universe is 156 billion light years wide or 1.5 × 1027 m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Q13: Obtain an expression for the frequency of radiation emitted when a hydrogen atom de excites from level n to level (n − 1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit. When a hydrogen atom de excites itself from a higher level to a lower level say from n to n – 1, find an expression for its frequency of radiation. Also for a large value of n, show that this frequency equals classical frequency of revolution of the electron in the orbit.

Ans:

It is given that an atom of hydrogen de excites from an upper level to a lower level (from n to n − 1).

We have the relation for energy (E1) of radiation at level n as:

E1=hv1=hme4(4π)3ϵ02(h2π)3×(1n2)E_1 = h v_1 = \frac{ h m e^{4} }{ (4 \pi )^3 { \epsilon_0 }^{2}( \frac{ h }{ 2 \pi } )^{ 3 }} \times \left ( \frac{ 1 }{ n^{2} } \right ) —–(1)

Where,

0 = Permittivity of free space

h = Planck’s constant = 6.63 × 10−34 Js

m = Mass of an atom of hydrogen

e = Charge of an electron = 1.9 × 10−19 C

v1 = Frequency of radiation at level n

Now, the relation for energy (E2) of radiation at level (n − 1) is given as:

E2=hv2=hme4(4π)3ϵ02(h2π)3×(1(n1)2)E_2 = h v_2 = \frac{ h m e^{4} }{ (4 \pi )^3 { \epsilon_0 }^{2}( \frac{ h }{ 2 \pi } )^{ 3 }} \times \left ( \frac{ 1 }{ (n-1)^{2} } \right )—–(2)

Where,

v2 = Frequency of radiation at level (n – 1)

Energy (E) released as a result of de-excitation:

E = E2E1

hν = E2E1—–(3)

Where,

ν = Frequency of radiation emitted

Putting values from equations (1) and (2) in equation (3), we get:

v=me4(4π)3ϵ02(h2π)3×(1(n1)21n2)v = \frac{ m e^{4} }{ (4 \pi )^3 { \epsilon_0 }^{2}( \frac{ h }{ 2 \pi } )^{ 3 }} \times \left ( \frac{ 1 }{ (n-1)^{2} }-\frac{1}{ n^{2} } \right ) v=me4(2n1)(4π)3ϵ02(h2π)3n2(n1)2v = \frac{ m e^{4} ( 2n – 1) }{ (4 \pi )^3 { \epsilon_0 }^{2}( \frac{ h }{ 2 \pi } )^{ 3 } n^{2} (n-1)^{2}}\\

For large n, we can write  ( 2n  – 1 ) \simeq 2n and ( n – 1 ) \simeq = n.

v=me432π3ϵ02(h2π)3n2v = \frac{ m e^{4} }{ 32 \pi^3 { \epsilon_0 }^{2}( \frac{ h }{ 2 \pi } )^{ 3 } n^{2} } —–(4)

Classical relation of frequency of revolution of an electron is:

vc=v/2πrv_c = v/2\pi r —–(5)

Where, Velocity of the electron in the nth orbit is:

v=e24πϵ0(h2π)nv=\frac{e^2}{4 \pi \epsilon_0 \left ( \frac{h}{2 \pi} \right ) n } —–(6)

And, radius of the nth orbit is:

r=4πϵ0(h2π)2me4n2r = \frac{ 4 \pi { \epsilon_0 }( \frac{ h }{ 2 \pi } )^{ 2 } }{ m e^{4} }n^{2}—–(7)

Putting the values of equations (6) and (7) in equation (v), we get:

v=me432π3ϵ02(h2π)3n3v = \frac{ m e ^ { 4 } } { 32 \pi ^ 3 { \epsilon_0 }^{ 2 }( \frac{ h }{ 2 \pi } ) ^ { 3 } n ^ { 3 } }

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

Q14: Normally, in an atom an electron can be expected to be in any orbit surrounding the nucleus. Then how do we calculate the size of an atom? Why is an atom not, say hundred times bigger than its actual size? This question had greatly troubled Bohr before he arrived at his model of an atom that has been said in the text. To get an idea of the simulations he must have to go through before his discovery, let us try as follows with the basic constants of nature and check if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10−10 m).

(i) Develop a value with the measurements of length from the key constants e, me, and C. Calculate its numerical value.

(ii) With your calculations, you will come to an answer wherein the value is much smaller than the actual size as it involves C. But we know that atomic energies are non-relativistic hence c should not play any part. This must have made Bohr to overlook c in search for a different factor to get the right atomic size. As Plank’s constant h had already made its appearance elsewhere, with its help Bohr was able to find the right atomic size. Find a quantity with dimension of length h, me, and e and verify that its numerical value has the correct order of magnitude.

Ans:

(a) Charge on an electron, e = 1.6 × 10−19 C

Mass of an electron, me = 9.1 × 10−31 kg

Speed of light, c = 3 ×108 m/s

Let the given quantities be: (e24πϵ0mec2)\left (\frac{e^2}{ 4 \pi \epsilon_0 m_e c^2} \right )

Where, ϵ0\epsilon_0  = Permittivity of free space

And, the numerical value of the taken quantity will be:

14πϵ0×e2mec2\frac{1}{4\pi\epsilon_0}\times\frac{e^2}{m_ec^2} =9×109×(1.6×1019)29.1×1031×(3×108)2= 9 \times 10^9 \times \frac { ( 1.6 \times 10^{ -19 } ) ^ 2 } { 9.1 \times 10^{ -31 } \times ( 3 \times 10^8 ) ^ 2 } = 2.81 × 10-15 m

Hence, the calculated value of size is much lower than the actual value of size of an atom.

(b) Charge on an electron, e = 1.6 × 10−19 C

Mass of an electron, me = 9.1 × 10−31 kg

Planck’s constant, h = 6.63 ×10−34 Js

Let us take a quantity involving the given quantities as 4πϵ0(h2π)2mee2\frac{4\pi\epsilon_0 \left ( \frac{h}{2\pi} \right )^{2}}{m_e e^2}

Where, ϵ0\epsilon_0 = Permittivity of free space

And, 14πϵ0\frac{1}{4 \pi \epsilon_0 } = 9 × 109 Nm2C-2

The numerical value of the taken quantity will be:

4πϵ0×(h2π)2mee24 \pi \epsilon_0 \times \frac{(\frac{h}{2\pi})^2}{m_e e^2} =19×109×(6.63×10342×3.14)29.1×1031×(1.6×1019)2= \frac { 1 }{ 9 \times 10^9} \times \frac{( \frac{ 6.63 \times 10^{ -34 }}{ 2 \times 3.14})^ 2}{9.1 \times 10^{ -31 } \times (1.6 \times 10^{-19})^2} = 0.53 × 10-19m

Hence, the value of the quantity taken is of the order of the atomic size.

Q15: In the case of a hydrogen atom, the energy of an electron (total energy) in the first excited state is -4.3 eV. In this state, find:

(a)  Kinetic energy of electron?

(b) Potential energy of electron?

(c) If we change the value of zero potential, which of the answers above would change?

Ans:

(a) Total energy of the electron, E = −4.3 eV

Kinetic energy of the electron is equal to the negative of the total energy.

K = −E = − ( 4.3 ) = +4.3 eV

Hence, the kinetic energy of the electron in the given state is +4.3 eV.

(b) Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.

U = −2 K = − 2 × 4.3 = − 8.6 eV

Hence, the potential energy of the electron in the given state is − 8.6 eV.

(c) According to the reference point taken potential energy varies.

The reference point of the potential energy is considered to be 0. The value of potential energy is dependent on the reference point. Therefore, any change in reference point will result in the change of potential energy which will bring changes in the total energy which is equal to the sum of potential energy and kinetic energy.

Q16: If Bohr’s quantization postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion as well. Why then do we never speak of quantization of orbits of planets around the sun?

Ans:

The quantum level for a planetary motion is considered to be continuous. This is because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). 1070h is the order of the angular momentum of the Earth in its orbit. As the values of n increase, the angular momenta decreases. So, the planetary motion is considered to be continuous.

Q17: 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−) of mass about 207me orbits around a proton].

Ans:

Mass of a negatively charged muon,  mP = 207 me

According to Bohr’s model:

Bohr radius, reα(1me)r_e \alpha \left ( \frac{1}{m_e} \right )

And, energy of a ground state electronic hydrogen atom, Ee    α  me\;\alpha\; m_e

Also, the energy of a ground state muonic hydrogen atom, Eu = mu

We have the value of the first Bohr orbit, re  = 0.53 A = 0.53 × 10-10 m

Let, rμ be the radius of muonic hydrogen atom.

Following is the relation at equilibrium:

mp rp = me re

207 me × ru = me re

Ru = ( 0.53 × 10-10 )/ 207 = 2.56 × 10-13 m

Hence, for a muonic hydrogen atom 2.56 × 10−13is the value of first Bohr radius.

We have,

Ee= − 13.6 eV

Considering the ratio of the energies we get:

EeEu=memu=me207  me\frac{E_e}{E_u}=\frac{m_e}{m_u}=\frac{m_e}{207\;m_e}\\

Eu = 207 Ee = 207 × ( – 13.6 ) = – 2.81 keV

−2.81 keV is the ground state energy of a muonic hydrogen atom.

 

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