CBSE Class 12 Physics Atoms MCQs Set F

Question: In the Bohr model of the hydrogen atom, the lowest orbit corresponds to
(a) infinite energy
(b) maximum energy
(c) minimum energy
(d) zero energy. 
Answer: c

Question: The binding energy of an electron in the ground state of He is equal to 24.6 eV. The energy required to remove both the electrons is
(a) 49.2 eV
(b) 54.4 eV
(c) 79 eV
(d) 108.8 eV 
Answer: c

Question: Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model?
(a) 0.65 eV
(b) 1.9 eV
(c) 11.1 eV
(d) 13.6 eV 
Answer: c

Question: Bohr’s basic idea of discrete energy levels in atoms and the process of emission of photons from the higher levels to lower levels was experimentally confirmed by experiments performed by
(a) Michelson–Morley
(b) Millikan
(c) Joule
(d) Franck and Hertz 
Answer: d

Question:(a) equal to one-fourth the circumference of the first orbit
(b) equal to half the circumference of first orbit
(c) equal to twice the circumference of first orbit
(d) equal to the circumference of the first orbit.
Answer: d

Question: If u₁ is the frequency of the series limit of Lyman series, u2 is the frequency of the first line of Lyman series and u₃ is the frequency of the series limit of the Balmer series, then
(a)v₁ –v₂ = u₃
(b) v₁ = v₂ – v₃
(c)1/ v₂=1/v₁+1/v₃ 
(d)1/v₁=1/ 1/v₂+1/v₃
Answer: a

Question: A hydrogen atom initially in the ground level absorbs a photon and is excited to n = 4 level then the wavelength of photon is
(a) 790 Å
(b) 870 Å
(c) 970 Å 
(d) 1070 Å 
Answer: c

Question: The wavelength limit present in the Pfund series is (R = 1.097 × 10⁷ m^–1)
(a) 1572 nm
(b) 1898 nm
(c) 2278 nm
(d) 2535 nm 
Answer: c

Question: The ratio of the speed of the electron in the ground state of hydrogen atom to the speed of light in vacuum is
(a) 1/2
(b) 2/237
(c) 1/137
(d) 1/237 
Answer: c

Question: The first line of the Lyman series in a hydrogen spectrum has a wavelength of 1210 Å.
The corresponding line of a hydrogen-like atom of Z = 11 is equal to
(a) 4000 Å
(b) 100 Å
(c) 40 Å
(d) 10 Å
Answer: d

Question: From quantisation of angular momentum, one gets for hydrogen atom, the radius of the nth orbit as r_n[n²/m_e][h/2π]²[4π²ε_o/e²]
For a hydrogen like atom of atomic number Z,
(a) the radius of the first orbit will be the same
(b) rn will be greater for larger Z values
(c) rn will be smaller for larger Z values
(d) none of these. 
Answer: c

Question: If muonic hydrogen atom is an atom in which a negatively charged muon (m) of mass about 207 me revolves around a proton, then first Bohr radius of this atom is (re = 0.53 × 10^–10 m)
(a) 2.56 × 10^–10 m
(b) 2.56 × 10^–11 m
(c) 2.56 × 10^–12 m
(d) 2.56 × 10^–13 m 
Answer: d

Question: The electric current I created by the electron in the ground state of H atom using Bohr model in terms of Bohr radius (a₀) and velocity of electron in first orbit v0 is
(a) ev₀/2πa_o
(b) 2πa/ev_o
(c) 2πa/v_o
(d) v_o/2πa
Answer: a

Question: If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å, the wavelength of the second line of the series should be
(a) 13122 Å
(b) 3280 Å
(c) 4860 Å
(d) 2187 Å
Answer: c

Question: If the radius of inner most electronic orbit of a hydrogen atom is 5.3 × 10–¹¹ m, then the radii of n = 2 orbit is
(a) 1.12 Å
(b) 2.12 Å
(c) 3.22 Å
(d) 4.54 Å 
Answer: b

Question: An electron is revolving in the n^th orbit of radius 4.2 Å, then the value of n is (r1 = 0.529 Å)
(a) 4
(b) 5
(c) 6
(d) 3 
Answer: d

Question: The excitation energy of Lyman last line is 
(a) the same as ionisation energy
(b) the same as the last absorption line in Lyman series
(c) both (a) and (b)
(c) different from (a) and (b) 
Answer: c

Assertion & Reasoning Based MCQs

Two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false

Question: Assertion (A) : According to classical theory, the proposed path of an electron in Rutherford atom model will be circular.
Reason (R) : According to electromagnetic theory an accelerated particle continuously emits radiation. 
Answer: b

Question: Assertion (A) : The force of repulsion between atomic nucleus and a-particle varies with distance according to inverse square law.
Reason (R) : Rutherford did a-particle scattering experiment.
Answer: b

Question: Assertion (A) : Electrons in the atom are held due to coulomb forces.
Reason (R) : The atom is stable only because the centripetal force due to Coulomb’s law is balanced by the centrifugal force.
Answer: c

Question: Assertion (A) : Total energy of revolving electron in any stationary orbit is negative.
Reason (R) : Energy is a scalar quantity. It can have positive or negative value.
Answer: b

Question: Assertion (A) : Between any two given energy levels, the number of absorption transitions is always less than the number of emission transitions.
Reason (R) : Absorption transitions start from the lowest energy level only and may end at any higher energy level. But emission transitions may start from any higher energy level and end at any energy level below it.
Answer: a

Question: Assertion (A) : For the scattering of a-particlesat large angles, only the nucleus of the atom is responsible.
Reason (R) : Nucleus is very heavy in comparison to electrons. 
Answer: a

Question: Assertion (A) : Balmer series lies in the visible region of electromagnetic spectrum. 
Reason (R) :  1/ λ=R[1/2²-1/K²,where K = 3, 4,5, …
Answer: b

Question: Assertion (A) : Fraunhofer lines are observed in the spectrum of the sun.
Reason (R) : The different elements have different spectra. 
Answer: b

Very Short Answer Type Questions 

Question: Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum does it belong.
Answer: The shortest wavelength of Brackett series is given as
1/λ=1.097×10⁷[1/4²-1/∞²]=1.097×10⁷/16.
⇒ λ = 1.4585 × 10–6 m
This wavelength lies in the infrared region of electromagnetic spectrum.

Question: Find the radius of the ground state orbit of hydrogen atom.
Answer: r_n=n²h²/4π²m_ekze²=n²/z(0.053nm)
For our case, n = 1 and Z = 2, and the result is r1 = 0.027 nm.

Short Answer Type Questions

Question: If Bohr’s quantisation postulate (angular momentum = nh/2π) 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 quantisation of orbits of planets around the sun?
Answer: Angular momentum mvr n=h /2π associated with planetary motion are incomparably large relative to h.
For example angular momentum of earth in its orbital motion is of the order of 10⁷⁰ h/2π .
For such large value of n, the difference in successive energies and angular momenta of the quantised levels of the Bohr model are so small that one can predict the energy level continuous.

Question: A small particle of mass m moves in such a way that the potential energy
U = ar² where a is a constant and r is the distance of the particle from the origin. Assuming Bohr’s model of quantisation of angular momentum for circular orbits, find the radius of nth allowed orbit.
Answer: The force at a distance r is
F =-dU/dr=− 2ar
Suppose r be the radius of nth orbit. The necessary centripetal force is provided by the above force. Thus,Further, the quantisation of angular momentum gives,mv²/r=2ar ..(i)
Further, the quantisation of angular momentum gives,
mvr=nh = 2π …(ii)

Solving, equations (i) and (ii) for r, we get r =[n²h²/8amπ ²]^1/4

Question: Find the number of unique radiations that can be emitted for a sample of hydrogen atoms excited to the nth level.
Answer: The first excited level is 2^nd line.
From the 2^nd level electron can go to level 1 ⇒ one radiation
3^rd level electron can go to levels 1, 2 ⇒ three radiations
4^th level electron can go to levels 1, 2, 3 ⇒ six radiations
n^th level electron can go to levels 1, 2, 3, …(n – 1)
∴ Total number of radiations
= 1 + 2 + …… + (n – 1) .n /2

Question: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?
Answer: Radius of innermost electron r=n²h²ε_o/πme²
For n = 1, r₁ = h²ε₀/πme²= 5.3 × 10^-11 m
For n = 2, r₂ = (2)² r₁ = 2.12 × 10^–10 m
For n = 3, r₃ = (3)² r₁ = 4.77 × 10^–10 m.

Long Answer Type Questions

Question: The first four spectral lines in the Lyman series of a H-atom are λ = 1218 Å, 1028 Å, 974.3 Å
and 951.4 Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
Answer: Let μ_H and μ_D are the reduced masses of electron for hydrogen and deuterium respectively.

Case Based MCQs
Second Postulate of Bohr’s Theory
Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niels Bohr, this electron moves in a stationary orbit. When this electron is in the stationary orbit, it emits no electromagnetic radiation. The angular momentum of the electron is quantized, i.e., mvr = (nh/2p), where m = mass of the electron, v = velocity of the electron in the orbit, r = radius of the orbit and n = 1, 2, 3, …. When transition takes place from Kth orbit to Jth orbit, energy photon is emitted. If the wavelength of the emitted photon is l, we find that 1/λ=R[1/j²-1/k²] , where R is Rydberg’s constant.
On a different planet, the hydrogen atom’s structure was somewhat different from ours. The angular momentum of electron was P = 2n(h/2p), i.e., an even multiple of (h/2π).

Question: In our world, the velocity of electron is v₀ when the hydrogen atom is in the ground state.
The velocity of electron in this state on the other planet should be
(a) v0
(b) v0/2
(c) v0/4
(d) v0/8
Answer: b

Question: In Bohr’s model of hydrogen atom, let PE represent potential energy and TE the total energy. In going to a higher orbit
(a) PE increases, TE decreases
(b) PE decreases, TE increases
(c) PE increases, TE increases
(d) PE decreases, TE decreases 
Answer: c

Question: Check the correctness of the following statements about the Bohr model of hydrogen atom.
(i) The acceleration of the electron in n = 2 orbit is more than that in n = 1 orbit.
(ii) The angular momentum of the electron in n = 2 orbit is more than that in n = 1 orbit.
(iii) The kinetic energy of the electron in n = 2 orbit is less than that in n = 1 orbit.
(a) Only (iii) and (i) are correct.
(b) Only (i) and (ii) are correct.
(c) Only (ii) and (iii) are correct.
(d) All the statements are correct.
Answer: c