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

Class 12 Physics NCERT book solutions for Chapter 12 - Atoms Questions and Answers.

Education Blogs Chapter 12 – Atoms Questions and Answers: NCERT Solutions for Class 12...

Class 12 Physics NCERT book solutions for Chapter 12 - Atoms Questions and Answers.

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).

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.

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

⇒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.

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 .

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.

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.

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.

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.

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.

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.

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 .

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?

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.

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.

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.

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.

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.

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.

1. Construct a quantity with the dimensions of length from the fundamental constants e,meandce,meandc. Also, determine its numerical value.

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,

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.

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.

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.

1. What is the kinetic energy of the electron in this state?

⇒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.

⇒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.

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.

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.

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.

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