CBSE Class 12 Physics Nuclei MCQs Set F

Question: If the nucleus 13Al²⁷ has a nuclear radius of about 3.6 fm, then ₅₂Te¹²⁵ would have its radius approximately as
(a) 9.6 fm
(b) 12 fm
(c) 4.8 fm
(d) 6 fm 
Answer: d

Question: Let m_p be the mass of a proton, mn the mass of a neutron, M₁ the mass of a ²⁰₁₀ Ne nucleus and M2 the mass of a ⁴⁰₂₀Ca nucleus. Then
(a) M₂ = M₁
(b) M₂ > 2M₁
(c) M₂ = 2M₁
(d) M₁ < 10 (m_n + m_p)
Answer: d

Question: The ratio of the nuclear radii of the gold isotope ¹⁹⁷₇₉Au and silver isotope ¹⁰⁷₄₇Ag is 
(a) 1.23
(b) 0.216
(c) 2.13
(d) 3.46 
Answer: a

 Question: The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of ⁴₂ He, if mass of ⁴₂He is 4.0015 u
(a) 0.0305 erg
(b) 0.0305 J
(c) 28.4 MeV
(d) 0.061 u 
Answer: c

Question: Light energy emitted by star is due to
(a) breaking of nuclei
(b) joining of nuclei
(c) burning of nuclei
(d) reflection of solar light 
Answer: b

 Question: Boron has two stable isotopes, ¹⁰₅B and ¹¹₅B.
Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of ¹¹₅B.
(a) 90.1%
(b) 80.1%
(c) 85.5%
(d) 95% B
Answer: b

 Question: 1 MeV positron encounters a 1 MeV electron travelling in opposite direction.
What is the wavelength of photons produced? (Given rest mass energy of electron or positron
= 0.512 MeV and h = 6.62 × 10^–34 Js)
(a) 8.2 × 10^–11 m
(b) 8.2 × 10^–13 m
(c) 8.2 × 10^–12 m
(d) 8.2 × 10^–9 m 
Answer: b

Question:Two stable isotopes ⁶₃Li and ⁷₃Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. The atomic weight of lithium is
(a) 6.941 u
(b) 3.321 u
(c) 2.561 u
(d) 0.621 u 
Answer: a

Question: The radius of a spherical nucleus as measured by electron scattering is 3.6 fm. What is the mass number of the nucleus most likely to be?
(a) 27
(b) 40
(c) 56
(d) 120
Answer: a

 Question: If in a nuclear fusion reaction, mass defect is 0.3%, then energy released in fusion of 1 kg mass is
(a) 27 × 10¹⁰ J
(b) 27 × 10¹¹ J
(c) 27 × 10¹⁰ J
(d) 27 × 10¹³ J 
Answer: d

Case Based MCQs

Nuclear Energy A heavy nucleus breaks into comparatively lighter nuclei which are more stable compared to the original heavy nucleus. When a heavy nucleus like uranium is bombarded by slow moving neutrons, it splits into two parts releasing large amount of energy. The typical fission reaction of 92U²³⁵.
92U²³⁵ + 0n¹ → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3₀n¹ + 200 MeV
The fission of 92U235 approximately released 200 MeV of energy.

Question: The release in energy in nuclear fission is consistent with the fact that uranium has
(a) more mass per nucleon than either of the two fragments
(b) more mass per nucleon as the two fragment 
(c) exactly the same mass per nucleon as the two fragments
(d) less mass per nucleon than either of two fragments.
Answer: a

Question: If 200 MeV energy is released in the fission of a single nucleus of ²³⁵ ₉₂U, the fissions which are required to produce a power of 10 kW is
(a) 3.125 × 10¹³
(b) 1.52 × 10⁶
(c) 3.125 × 10¹²
(d) 3.125 × 10¹⁴
Answer: d

Question: A nuclear fission is said to be critical when multiplication factor or K
(a) K = 1
(b) K > 1
(c) K < 1
(d) K = 0 
Answer: a

Question: When 92U235 undergoes fission, about 0.1% of the original mass is converted into energy. The energy released when 1 kg of 92U235 undergoes fission is
(a) 9 × 10¹¹ J
(b) 9 × 10¹³ J
(c) 9 × 10¹⁵ J
(d) 9 × 10¹⁸ J
Answer: b

Question: Einstein’s mass-energy conversion relation
E = mc² is illustrated by
(a) nuclear fission
(b) atomic transition
(c) rocket propulsion
(d) steam engine 
Answer: a

Discovery of Nucleus The nucleus was first discovered in 1911 by Lord Rutherford and his associates by experiments on scattering of a-particles by atoms. He found that the scattering results could be explained, if atoms consist of a small, central, massive and positive core surrounded by orbiting electrons.
The experimental results indicated that the size of the nucleus is of the order of 10–14 m and is thus 10000 times smaller than the size of atom.

Question: Masses of nuclei of hydrogen, deuterium and tritium are in ratio
(a) 1 : 2 : 3
(b) 1 : 1 : 1
(c) 1 : 1 : 2
(d) 1 : 2 : 4
Answer: a

Question: Ratio of mass of nucleus with mass of atom isv approximately
(a) 1
(b) 10
(c) 10³
(d) 10¹⁰
Answer: a

Question: If R is the radius and A is the mass number, then log R versus log A graph will be
(a) a straight line
(b) a parabola
(c) an ellipse
(d) none of these. 
Answer: a

Question: The ratio of the nuclear radii of the mercury isotope 19880Hg and silver isotope 107 47Ag is
(a) 1.23
(b) 0.216
(c) 2.13
(d) 3.46 
Answer: a

Question: Nuclides with same neutron number but different atomic number are
(a) isobars
(b) isotopes
(c) isotones
(d) none of these
Answer: c

Assertion & Reasoning Based MCQs
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) : For the fission of heavy nuclei, neutrons are more effective than protons.
Reason (R) : Neutrons are heavier than protons. 
Answer: b

Question: Assertion (A) : Energy is released in a nuclear reaction.
Reason (R) : In any nuclear reaction the reactants and resultant products obey the law of conservation of charge and mass only. 
Answer: c

Question: Assertion (A) : Two protons can attract each other.
Reason (R) : The distance between the protons within the nucleus is about 10–15 m. 
Answer: a

Question: Assertion (A) : The nuclear force becomes weak if the nucleus contains too many protons compared to neutrons.
Reason (R) : The electrostatic forces weaken the nuclear force.
Answer: c

Question: Assertion (A) : There is a chain reaction when uranium is bombarded with slow neutrons.
Reason (R) : When uranium is bombarded with slow neutrons more neutrons are produced. 
Answer: a

Very Short Answer Type Questions 

Question: A neutron is absorbed by a 63 Li nucleus with the subsequent emission of an alpha particle. Calculate the energy released, in MeV, in this reaction.
[Given : mass ⁶₃ Li = 6.015126 u;
mass (neutron) = 1.0086654 u;
mass (alpha particle) = 4.0026044 u;

Answer: ₃Li⁶ + ₀n¹ → ₂He⁴ + ₁H³ + Q
Total initial mass = 6.015126 + 1.0086654 = 7.0237914 amu
Total final mass = 4.0026044 + 3.01 = 7.0126044 amu
Mass defect, Dm = 7.0237914 – 7.0126044 = 0.0111870 amu
Energy released, Q = 0.0111870 × 931 = 10.415 MeV.

Short Answer Type Questions 

Question: 1 ₉₂U²³⁵ absorbs a slow neutron (thermal neutron) and undergoes a fission represented by
₉₂U²³⁵ + 0n¹ → ₉₂^U236 → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3₀n¹ + E.
Calculate :
(i) The energy released E per fission.
(ii) The energy released when 1 g of 92U235
undergoes complete fission.
Given : ₉₂U²³⁵ = 235.1175 amu (atom),
₅₆Ba¹⁴¹ = 140.9577 amu (atom)
₃₆Kr⁹² = 91.9264 amu (atom), 0n1 = 1.00868 amu,
1 amu = 931 MeV/c2

Answer: (i) E = [M_U + m_n – M_Ba – M_Kr – 3m_n] × 931 = 200.57 MeV
(ii) Energy released =N_A /235XE
=200.57X10⁶X1.6X10^-19X6.023X10²³/235X3.6X10⁶=22.84MWh

Question: Find the disintegration energy Q for the fission event represented by equation
92^U235 + 0n¹ → 92^U236 → 140₅₈Ce + ⁹⁴₄₀Zr + 2₀n¹
If mass of ₉₂U²³⁵ = 235.0439 u, ₀n¹ = 1.00867 u,
140₅₈Ce = 139.9054 u and ⁹⁴₄₀Zr = 93.9063 u, find energy released in the process.
Answer: The mass lost in the process,
Dm = 235.0439 + 1.00867 – (139.9054 + 93.9063 + 2.01734) = 0.22353 u
The corresponding energy released
= Δmc² = 0.22353 × 931 MeV = 208 MeV 

Question: If both the number of protons and neutrons in a nuclear reaction is conserved, in what way is mass converted into energy (or vice versa)?
Explain giving one example.
Answer: A certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy DEb will be released in this process.
The energy DEb is called the binding energy of the nucleus.
If we separate a nucleus into its nucleons we would have to transfer a total energy equal to DEb, to the nucleons.
Example : ₉₂U²³⁵+₀n¹→¹⁴¹₅₆ Ba+⁹²₃₆ Kr + 3¹₀ n + Q
The energy (Q) released was estimated to be 200 MeV per fission and is equivalent to the difference in masses of the nuclei before and after the fission.

Question: Calculate the energy released in fusion reaction : 
²₁H+²₁H→³₂He+n,where B.E. of ²₁H = 2.23 MeV and of ³₂ He = 7.73 MeV.
Answer: Given fusion reaction,
²₁H+²₁H→³₂He+n,
Energy released = final B.E. – initial B.E.
= 7.73 – (2.23 + 2.23) = 3.27 MeV.