Q: A positive ion having just one electron ejects it if a photon of wavelength 228A˚ or less is absorbed by it. Identifying the ion.

Given, λ=228A∘ E=λhc =228×10−106.63×10−34×3×108 =0.0972×10−16 The transition takes place form n=1 to n=2 Now, 1.6×10−19×13.6×43×Z2 =0.0872×10−16 Z2=13.6×3×16×10−190.0872×10−16×4=5.3 Z=5.3=2.3 The ion may be Helium.

Q: hydrogen atom in n=6 makes two successive transitions & reaches the ground state. In the first transition a photon of 1.13eV is emitted. Find the energy of the photon emitted in the second transition \& value of n for the intermediate state.

n=6,E=36−13.6=−0.3777777 Energy in grounds tate =−13.6eV. Energy emitted in Second transition =−13.6−(0.37777+1.13) =−12.09=12.1eV (b) Energy in the intermediate state =1.13eV+0.37 implies n218.6×Z2=n213.0=15077 or n=1.50713.6=3.03=3

Q: Find the quantum number n corresponding to the excited state of He+ion if on transition to the ground state that ion emits two photons in succession with wave lengths 108.5 and 30.4 nm

Given λ2=30.4×10−7 cm λ1=108.5×10−7 cm Suppose excited state of He+is n2. The electron falls from n1 to n1 and then n1 to 1 energy level to emit two successive photons. ∴λ21=RH×Z2[121−n121] or 30.4×10−71=109678×4[121−n121] ∴n1=2 Now for λ1:n1=2 and n2=n ∴λ11=RH×Z2[221−n31] or 108.5×10−71=109678×4[221−n221] Thus, the excited state of He+is 5 th orbit.

Q: (a) Find the maximum wavelength λ of light which can ionize a H-atom in ground state. (b) Light of wavelength λ incident on a H-atom which is in its first excited state. Find the kinetic energy of the electron coming out.

(a) En=n213.6 (for H-atom) E=13.6eV (Ground state) Ionisation energy =13.6eV Max. wave length req. 13.6×1.6×10−19=λmaxhc=λmax6.6×10−34×3×108⇒λmax913A˚ (b) First excited state n=2 total energy of electron ∴En=2=413.6eV=−3.4eV ∴ So work function ϕ=3.4eV maximum Kinetic energy = Energy incident-work function ⇒13.6−3.4=10.2eV

Q: Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2 . (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm)

(a) n1=1,n2=3 E=13.6(11−91)=13.6×98=λhc or 913.6×8=λ4.14×10−15×3×108 ⇒λ=13.6×84.14×10−7×3=103 nm (b) As n changes by 2 , we may consider n=2 to n=4 then E=13.6(41−161)=13.6×6416−4=2.55eV and 2.55=λ1242 or λ=2.551242=487 nm

Q: A doubly ionized Lithium atom is hydrogen - like with atomic number 3; (a) Find the wavelength of radiation required to excite the electron in L++from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6eV ). (b) How many spectral lines are observed in the emission spectrum of the above excited system?

(a)In hydrogen atom, ΔE1=E3−E1=12.1eV E∝Z2 ∴F or Z=3 ΔE2=((3))2ΔE1=((3))2((12.1)) =108.9A˚ ∴Λ=108.912375≈113A˚ (b) The possible lines in the emission spectrum of this excited system are [2nn−1=323−1=3]

Q: A parallel beam, of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon moving in the same direction as the incident photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.

λ=100 nm E=λhc=1001242=12.42eV a) The possible transitions may be E1 to E2 E1 to E2, energy absorbed =10.2eV. Energy left =12.42−10.2=2.22eV. Energy left =12.42−10.2=2.22eV. 2.22eV=λhc=λ1242 orλ=559.45=560 nm E1 to E3, Energy absorbed =12.1eV Energy left =12.42−12.1=0.32eV 0.32=λhe=λ1242 orλ=0.321242=3881.2=3881 nm E3 to E4, Energy absorbed =0.65 Energy left =12⋅42−0.65=11.77eV 11.77=λhc=λ1242 orλ=11.771242=105.52 b)The energy absorbed by the H atom is now radiated perpendicular to the incident beam. →10.2=λhc or λ=10.21242=121.76 nm →12.1=λhcorλ=12.11242=102.64 nm →0.65=λhcorλ=0.651242=1910.76 nm

Question 1

A positive ion having just one electron ejects it if a photon of wavelength 228A˚ or less is absorbed by it. Identifying the ion.

Accepted Answer

Given, λ=228A∘
E=λhc​
=228×10−106.63×10−34×3×108​
=0.0972×10−16
The transition takes place form 
n=1 to n=2
Now, 1.6×10−19×13.6×43​×Z2 
=0.0872×10−16
Z2=13.6×3×16×10−190.0872×10−16×4​=5.3
Z=5.3​=2.3
The ion may be Helium.

Question 2

A beam of ultraviolet radiation having wavelength between 100 nm and 200 nm is incident on a sample of atomic hydrogen gas. Assuming that the atoms are in ground state, which wavelengths will have low intensity in the transmitted beam? If the energy of a photon is equal to the difference between the energies of an excited state and the ground state, it has large probability of being absorbed by an atom in the ground state.

Accepted Answer

The energy of a photon corresponding to λ=100 nm is
100 nm1242eVnm​=12.42eV
and that corresponding to λ=200 nmis6.21eV
The energy needed to take the atom from the ground state to the first excited state is
E2​−E1​=13.6eV−3.4eV=10.2eV
To the second excited state is
E2​−E1​=13.6eV−1.5eV=12.1eV
To the third excited state is
E2​−E1​=13.6eV−0.85eV=12.75eV
At 10.2eV
Wavelength 1=10.21242​=122 nm
At 12.1 eV
Wavelength 2=12.11242​=103 nm
These wavelength will have low intensity in the tranmitted beam

Question 3

hydrogen atom in n=6 makes two successive transitions & reaches the ground state. In the first transition a photon of 1.13eV is emitted. Find the energy of the photon emitted in the second transition \& value of n for the intermediate state.

Accepted Answer

n=6,E=36−13.6​=−0.3777777
Energy in grounds tate =−13.6eV.
Energy emitted in Second transition
=−13.6−(0.37777+1.13)
=−12.09=12.1eV
(b) Energy in the intermediate state
=1.13eV+0.37
implies n218.6×Z2​=n213.0​=15077
or n=1.50713.6​​=3.03=3

Question 4

Find the quantum number n corresponding to the excited state of He+ion if on transition to the ground state that ion emits two photons in succession with wave lengths 108.5 and 30.4 nm

Accepted Answer

Given λ2​=30.4×10−7 cm
λ1​=108.5×10−7 cm
Suppose excited state of He+is n2​. The electron falls from n1​ to n1​ and then n1​ to 1 energy level to emit two successive photons.
∴λ2​1​=RH​×Z2[121​−n12​1​]
or 30.4×10−71​=109678×4[121​−n12​1​]
∴n1​=2
Now for λ1​:n1​=2 and n2​=n
∴λ1​1​=RH​×Z2[221​−n31​]
or 108.5×10−71​=109678×4[221​−n22​1​]
Thus, the excited state of He+is 5 th orbit.

Question 5

(a) Find the maximum wavelength λ of light which can ionize a H-atom in ground state.
(b) Light of wavelength λ incident on a H-atom which is in its first excited state. Find the kinetic energy of the electron coming out.

Accepted Answer

(a) En​=n213.6​ (for H-atom) 
E=13.6eV (Ground state) 
Ionisation energy =13.6eV 
Max. wave length req.
13.6×1.6×10−19=λmax​hc​=λmax​6.6×10−34×3×108​⇒λmax​913A˚
(b) First excited state n=2 
total energy of electron 
∴En=2​=413.6​eV=−3.4eV
∴ So work function ϕ=3.4eV
maximum Kinetic energy = Energy incident-work function 
⇒13.6−3.4=10.2eV

Question 6

A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 500A˚. Assuming that the entire photon energy is takes up by the electrons, with what kinetic energy will the electron be ejected.

Accepted Answer

λ=50 nm
Work function = Energy required to remove the electron from n1​=1 to n2​=∞
E=13.6(1/1−1/∞)=13.6
λhc​−ϕ=KE
λhc​−13.6=KE
⇒501242​−13.6=KE⇒KE=24.84−13.6=11.24eV.

Question 7

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2 .
(a) Find the smallest wavelength emitted by hydrogen.
(b) List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm)

Accepted Answer

(a) n1​=1,n2​=3
E=13.6(11​−91​)=13.6×98​=λhc​
or 913.6×8​=λ4.14×10−15×3×108​
⇒λ=13.6×84.14×10−7×3​=103 nm
(b) As n changes by 2 , we may consider n=2 to n=4 
then E=13.6(41​−161​)=13.6×6416−4​=2.55eV 
and 2.55=λ1242​ or λ=2.551242​=487 nm

Question 8

A particle of mass m moves along a circular orbit in a centrosymmetrical potential field U(r)=2kr2​. Using the Bohr's quantization condition, find the permissible orbital radii and energy levels of that particle.

Accepted Answer

We have the following equation (we ignore reduced mass effects), thus we know that
rmv2​=kr
also, mvr=nh
so, v=mk​ r
and r=mk​nℏ​​ and v=nhmk​​/m
The energy levels are En​=21​ mv2+21​krr2
=21​mn−mk​​+21​kmk​nh​
=nπmk​​

Question 9

A doubly ionized Lithium atom is hydrogen - like with atomic number 3;
(a) Find the wavelength of radiation required to excite the electron in L++from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6eV ).
(b) How many spectral lines are observed in the emission spectrum of the above excited system?

Accepted Answer

(a)In hydrogen atom,
ΔE1​=E3​−E1​=12.1eV
E∝Z2
∴F or Z=3
ΔE2​=((3))2ΔE1​=((3))2((12.1))
=108.9A˚
∴Λ=108.912375​≈113A˚
(b) The possible lines in the emission spectrum of this excited system are [2nn−1​=323−1​=3]

Question 10

A parallel beam, of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state.
(a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon moving in the same direction as the incident photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam?
(b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.

Accepted Answer

λ=100 nm
E=λhc​=1001242​=12.42eV
a) The possible transitions may be 
E1​ to E2​ E1​ to E2​, energy absorbed =10.2eV.
Energy left =12.42−10.2=2.22eV.
Energy left =12.42−10.2=2.22eV.
2.22eV=λhc​=λ1242​
orλ=559.45=560 nm
E1​ to E3​, Energy absorbed =12.1eV
Energy left =12.42−12.1=0.32eV
0.32=λhe​=λ1242​
orλ=0.321242​=3881.2=3881 nm
E3​ to E4​, Energy absorbed =0.65
Energy left =12⋅42−0.65=11.77eV
11.77=λhc​=λ1242​
orλ=11.771242​=105.52
b)The energy absorbed by the H atom is now radiated perpendicular to the incident beam.
→10.2=λhc​ or λ=10.21242​=121.76 nm
→12.1=λhc​orλ=12.11242​=102.64 nm
→0.65=λhc​orλ=0.651242​=1910.76 nm

Question 11

A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204eV. If it makes a transition to quantum state n, a photon of energy 40.8eV is emitted. Find n,Z and the ground state energy (in eV ) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is −13.6eV.

Accepted Answer

Let ground state energy (in eV) be E1​ 
Then from the given condition 
E2n​−E1​=204eV
or 4n2E1​​−E1​=204eV
or E1​(4n21​−1))=204eV ....(i) 
or E2n​−En​=40.8eV 
or 4n2E1​​−n2E1​​=40.8eV
or E1​(4n2−3​)=40.8eV ....(ii)
From Eqs. (i) and (ii), we get
4n23​1−4n21​​=5or1=4n21​+4n215​
or n24​=1 or n=2
From Eq. (ii) E1​=−34​n2(40.8)eV
=−34​n2(40.8)eV
or E1​=−217.6eV
E1​=−(13.6)Z2
∴Z2=13.6E1​​=−13.6−217.6​=16
∴Z=4
Emin​=E2n​−E2n−1​=4n2E1​​−((2n−1))2E1​​)
∴Emin​=10.58eV

Question 12

A stationary He+ion emitted a photon corresponding to the first line of the Lyman series. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. Find the velocity of photoelectron.

Accepted Answer

Photon energy liberated from He+=(he/λ)
In Itst line of Lyman series =hH​×Z2[(121​−221​)]
=6.626×10−27×3.0×1010×109678×22×[3/4]
=6.54×10−11erg
This energy is used in liberating electron from H-atom from the ground state and therefore
6.54×10−11=E1​ of H+KE of electron given out [EH​=2.18×10−11erg]
∴21​mu2=4.362×10−11
or u2=9.108×10−284.362×10−11×2​
∴u=3.09×108cmsec−1

Question 13

The average kinetic energy of molecules in a gas at temperature T is 1.5KT. Find the temperature at which the average Kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular form at this temperature?

Accepted Answer

KE=3/2KT=1.5KT
K=8.62×10−5ev/k
Binding energy =−13.6(1/∞−1/1)=13.6eV
According to the question, 1.5KT=13.6eV
⇒1.5×8.62×10−5×T=13.6
⇒T=1.5×8.62×10−513.6​=1.05×105 K
No, because the molecule exists an H2+​which is impossible.

Question 14

A well collimated parallel pencil of cathode rays falls through a potential difference 3kV& enters the spacing between two parallel metallic plates, parallel to their length the spacing between the plates being 0.5 cm. The pencil strikes a fluorescent screen, mounted perpendicular to the length of the plates at the farther end of the plates & produces fluorescent spot. If now a potential difference of 3 V is applied across the two plates, calculate the linear deflection of the spot on the screen. Given the length of the plates is 10 cm.

Accepted Answer

[0.05 cm]

Question 15

A hydrogen in a state having a binding energy of 0.85eV makes transition to a state with excitation energy is 10.2eV.
(a) Identify the quantum no. n of an upper and the lower energy states involved in the transition.
(b) Find the wavelength of the emitted radiation.

Accepted Answer

[4,2,4870A˚]

Question 16

A hypothetical, hydrogen like atom consists of a nucleus of mass m1​ and charge (+Ze) and a mu-meson of mass m2​ and charge (−e). Using Bohr's theory, derive an expression for distance between nucleus and mu-meson for principal quantum number n and derive a relation for energy also. Hence obtain expression for reduced mass.

Accepted Answer

[r=πm1​m2​Ze2ε0​n2h2(m1​+m2​)​,E=−81​⋅(m1​+m2​)ε02​n2h2m1​m2​Z2e4​,m0​=m1​+m2​m1​m2​​]

Question 17

Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does latter photon correspond to ?

Accepted Answer

Lyman followed by Balmer i.e from n=2 to n=1
n1​=2 to n2​=1
λ1​=R(n22​1​−n22​1​)⟹λ1​=1.097×107(221​−121​)
⟹1.097×107(41​−1)
⟹λ=1.097×34​×10−7
=1.215×10−7=121.5×10−9=122 nm

Question 18

A particular hydrogen-like ion emits radiation of frequency 2.467×1015 Hz when it makes transition from n=2 to n=1. What will be the frequency of the radiation emitted a transition from n=3 to n=1?

Accepted Answer

The frequency of radiation emitted is given by v=λς​=K(n12​1​−n22​1​)
Thus 2.467×1015 Hz=K(121​−221​) 
or K=34​×2.467×1015 Hz
The frequency of the radiation emitted in the transition 
n=3→n=3 is
v1=K[121​−321​]
=98​K=98​2.467×1015 Hz
=2.92×1015 Hz.

Question 19

Monochromatic radiation of wavelength λ is incident on hydrogen sample in ground state. H-atom absorbs a fraction of light \& subsequently emit radiation of six different wavelengths. Find the value of λ.

Accepted Answer

As the hydrogen atoms emit radiation of six difference wavelength some of then must have been excited to n=4 The energy n=4 state is
E0​=42E1​​=−1613.6eV​=−0.85eV
Theenergy needed to take a hydrogen atom from its ground stateto n=4 is 
13.6eV−0.85eV=12.75eV
The photons of the incident rediation have 12.75eV of energy So
λhc​=12.75eV
or , lambda =(hc)/(12.75eV)=(1242evnm)/(12.75eV)=97.5 nm2

Question 20

A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of the electronic charge. It requires 47.2eV to excite the electron from the second Bohr orbit to the third Bohr orbit. Find :

Accepted Answer

This atom is hydrogen like
Z = atomic number of the nucleus
En​= Energy of the electron in the nth orbit =(Z)2 (energy of the electron in the nth orbit of the hydrogen atom)
=−(Z)213.6/n2eV=E1​/n2
where E1​= Energy of the electron in the 1st Bohr orbit of the given atom.
(a) Given (Z2)(13.6)(221​−321​)
=47.2eV
⇒Z=5
(b) E1​=−(25)113.6​=−340eV
E4​−E3​=13.6Z2[421​−321​]
=E1​[321​−421​]=340[321​−421​]eV
=16.53eV.
(c) minimum energy required to remove electron from first orbit = 340eV
⇒λhc​=340×1.6×10−19
or λ=[340×1.6×10−196.6×10−34×3×108​]m
=36.40A˚
(d) KE of the electron in the 1st orbit
KE1​=−E1​=340eV;PE1​=2E1​=−680eV
Angular momentum of the electron in the 1 st Bohr orbit
=2πh​=2π6.63×10−34​
=1.055×10−34 kg m2 s−1
(e) Radius of the 1 st Bohr orbit for the given atom
=ZBohrradius​=55.3×10−11​
=1.06×10−11 m

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Solutions for Physics Galaxy Optics and Modern Physics Vol 4 by Ashish Arora

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