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JEE • PHYSICS • NUCLEI

Nuclei practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 351 questions
1 Medium
1 of 351
Application +4 | -1
The radius of a nucleus with atomic mass number 7\mathrm{7} is 2  fermi\mathrm{2}\,\,\mathrm{fermi}. Find the radius of nucleus with atomic number 189\mathrm{189}.
View answer and explanation
Correct option: D
6  fermi\mathrm{6}\,\,\mathrm{fermi}
Option D (6 fermi) is correct because the nuclear radius is proportional to the cube root of the mass number.
2 Medium
2 of 351
Understanding +4 | -1
Which one is correct about fission?
View answer and explanation
Correct option: A
Approx. 0.1%0.1 \% mass converts into energy
Option A (Approx. 0.1%0.1 \% mass converts into energy) is correct because in nuclear fission, a small fraction of the mass is converted into energy according to Einstein's mass-energy equivalence.
3 Medium
3 of 351
Application +4 | -1
Ne22N{e^{22}} nucleus, after absorbing energy, decays into two α−\alpha - particles and an unknown nucleus. The unknown nucleus is
View answer and explanation
Correct option: B
Carbon
Option B (Carbon) is correct because the atomic number and mass number balance in the nuclear reaction.
4 Medium
4 of 351
Understanding +4 | -1
A moderator is used in nuclear reactors in order to
View answer and explanation
Correct option: A
Slow down the speed of the neutrons
The correct answer is A (Slow down the speed of the neutrons) because moderators are used to reduce the kinetic energy of neutrons to increase the probability of nuclear fission.
5 Medium
5 of 351
Remembering +4 | -1
Nuclear forces are
View answer and explanation
Correct option: D
All the above
Option D (All the above) is correct because nuclear forces exhibit all the listed properties: they are non-central, saturated, and spin-dependent.
6 Medium
6 of 351
Understanding +4 | -1
For uranium nucleus how does its mass vary with volume?
View answer and explanation
Correct option: A
m∝Vm \propto V
Option A (m∝Vm \propto V) is correct because nuclear density is constant, implying mass is directly proportional to volume.
7 Medium
7 of 351
Application +4 | -1
The half-life of a radioactive isotope ‘x’ is 60 years. It decays to another element ‘y’ which is stable. Two elements x and y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be
View answer and explanation
Correct option: D
240 years
Option A (5 m) is correct because the distance is calculated by finding the time it takes for the box to fall off the trolley and then using the lorry's acceleration to find the distance covered.
8 Medium
8 of 351
Understanding +4 | -1
Statement-I: The fusion process is achived by raising the temperature of the system.

Statement-II: The fusion particles have enough K.E to overcome the coulomb repulsive behaviour.
View answer and explanation
Correct option: A
Statement-I and Statement-II are true
Option A is correct because both statements are true: fusion requires high temperatures to overcome Coulomb repulsion.
9 Medium
9 of 351
Application +4 | -1
Two radioactive elements AA and BB initially have same number of atoms. The half life of AA is same as the average life of BB. If λA\lambda_A and λB\lambda_B are decay constants of AA and BB respectively, then choose the correct relation from the given options.
View answer and explanation
Correct option: C
λA=λBln⁡2\lambda_A=\lambda_B \ln 2
Option C (λA=λBln⁡2\lambda_A=\lambda_B \ln 2) is correct because it relates the decay constants based on the given half-life and average life relationship.
10 Medium
10 of 351
Application +4 | -1
In an α−\alpha - decay, the kinetic energy of α−\alpha - particle is 48 MeV and Q value of the reaction is 50 MeV. The mass number of the mother nucleus is [assume that daughter nucleus is in ground state]
View answer and explanation
Correct option: B
100
Option B (100) is correct because the mass number of the mother nucleus can be calculated using the kinetic energy of the alpha particle and the Q value of the reaction.
11 Medium
11 of 351
Understanding +4 | -1
A free neutron decays into a proton, an electron and
View answer and explanation
Correct option: B
An antineutrino
Option B (An antineutrino) is correct because a free neutron decays into a proton, an electron, and an antineutrino to conserve lepton number.
12 Medium
12 of 351
Analyzing +4 | -1
Assertion: A free neutron decays to a proton but a free proton
does not decay to a neutron.
Reason: Neutron has larger rest mass than the proton.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D is correct. A free neutron decays into a proton, an electron, and an antineutrino. A free proton cannot decay into a neutron because it violates energy conservation. The neutron has a larger rest mass than the proton, so the reason is correct, but the assertion is false.
13 Medium
13 of 351
Application +4 | -1
In order to decrease radioactive nuclei to one million of its initial number, number of half – lives required is
View answer and explanation
Correct option: A
20
Option A (20) is correct because it takes 20 half-lives to reduce a radioactive sample to one millionth of its original amount.
14 Medium
14 of 351
Understanding +4 | -1
Statement-I: Nuclear forces arise from strong coulombic interactions between protons and neutrons.

Statement-II: Nuclear forces are independent of charge of the nucleons.
View answer and explanation
Correct option: D
Statement-I is false and Statement-II is true
Option D is correct because nuclear forces are not Coulombic and are independent of the charge of the nucleons.
15 Medium
15 of 351
Remembering +4 | -1
The element used for radioactive carbon dating for more than 56000 yr\mathrm{56000 }\,\mathrm{yr} is
View answer and explanation
Correct option: A
C−14\mathrm{C - 14}
Option A (C−14\mathrm{C - 14}) is correct because Carbon-14 is the isotope used for radioactive dating of organic materials up to around 50,000 years.

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16 Medium
16 of 351
Remembering +4 | -1
If MM is the atomic mass and AA is the mass number, packing fraction is given by
View answer and explanation
Correct option: B
M−AA\frac{{M - A}}{A}
The correct answer is B (M−AA\frac{{M - A}}{A}) because packing fraction is defined as the ratio of mass excess to the mass number.
17 Medium
17 of 351
Application +4 | -1
The radius RR of a nucleus of mass number AA can be estimated by the formula R=(1.3×10−15)A1/3m.R = (1.3 \times \mathrm{1}{\mathrm{0}^{ - 15}}){A^{1/3}}m. It follows that the mass density of a nucleus is of the order of (Mproton≅Mneutron≃1.67×10−27  kg)({M_{proton}} \cong {M_{neutron}} \simeq 1.67 \times {10^{ - 27}}\mathrm{\,\,kg)}
View answer and explanation
Correct option: A
1017  kg m−3{10^{17}}\mathrm{\,\,kg}\,{\mathrm{m}^{ - 3}}
Option A (1017 kg m−3{10^{17}}\mathrm{\ kg}\,{\mathrm{m}^{ - 3}}) is correct because it is the approximate mass density of a nucleus, independent of the mass number A.
18 Medium
18 of 351
Remembering +4 | -1
Curie is a unit of
View answer and explanation
Correct option: C
Radioactivity
Option C (Radioactivity) is correct because Curie is the unit used to measure the radioactivity of a substance.
19 Medium
19 of 351
Remembering +4 | -1
According to law of conservation of energy and momentum for β-decay, the existence of a new particle along with electron was predicted, which was
View answer and explanation
Correct option: D
neutrino
Option D (neutrino) is correct because the neutrino's existence was predicted to conserve energy and momentum in beta decay.
20 Medium
20 of 351
Analyzing +4 | -1
Given the masses of various atomic particles mp=1.0072 u,    mn=1.0087  u,    me=0.000548  u,    mvˉ=0,    md=2.0141  u,{m_p} = 1.0072\,u,\,\,\,\,{m_n} = 1.0087\,\,u,\,\,\,\,{m_e} = 0.000548\,\,u,\,\,\,\,{m_{\bar v}} = 0,\,\,\,\,{m_d} = 2.0141\,\,u, Where pp = proton, nn = neutron, ee = electron vˉ=\bar v = antineutrino and dd = deuteron. Which of the following process is allowed by momentum and energy conservation?
View answer and explanation
Correct option: C
n+p→d+vˉn + p \to d + \bar v
Option B (p→n+e++vˉp \to n + {e^ + } + \bar v) is correct because the mass of the proton is less than the combined mass of the neutron, positron, and antineutrino, thus violating energy conservation. Option D (n+p→d+vˉn + p \to d + \bar v) is correct because the mass on the reactant side is greater than the product side.
21 Medium
21 of 351
Understanding +4 | -1
The mass density of a nucleus varies with mass number A\mathrm{A} as
View answer and explanation
Correct option: C
A0\mathrm{A}^{0}
The correct answer is C (A⁰) because the mass density of a nucleus is approximately constant and independent of the mass number A.
22 Medium
22 of 351
Remembering +4 | -1
The penetrating powers of α,  β  and  γ\mathrm{\alpha },\,\,\mathrm{\beta }\,\,\mathrm{and}\,\,\mathrm{\gamma } radiations, in decreasing order are
View answer and explanation
Correct option: B
γ,  β,  α\mathrm{\gamma },\,\,\mathrm{\beta },\,\,\mathrm{\alpha }
Option B (γ, β, α\rm{{\gamma}},\,\rm{{\beta}},\,\rm{{\alpha }}) is correct because gamma rays have the highest penetrating power, followed by beta particles, and then alpha particles.
23 Medium
23 of 351
Understanding +4 | -1
The correct relation between tav=t_{\mathrm{av}}= average life and t1/2=t_{1 / 2}= half life for a radioactive nuclei.
View answer and explanation
Correct option: C
0.693tav=t1/20.693 t_{\mathrm{av}}=t_{1 / 2}
Option C (0.693tav=t1/20.693 t_{\mathrm{av}}=t_{1 / 2}) is correct because it represents the relationship between average life and half-life derived from radioactive decay equations.
24 Medium
24 of 351
Understanding +4 | -1
A photon of 1.7×10−13  Joules1.7 \times {10^{ - 13}}\;Joules is absorbed by a material under special circumference. The correct statement is
View answer and explanation
Correct option: B
Electron and positron pair will be created
Option B (Electron and positron pair will be created) is correct because the energy of the photon is greater than 1.02 MeV, which is the minimum energy required for pair production.
25 Medium
25 of 351
Application +4 | -1
How many neutrons are more than protons in 92U235_{92}{\mathrm{U}^{235}} nucleus?
View answer and explanation
Correct option: C
51
Option C (51) is correct because it is the difference between the number of neutrons and protons in the Uranium nucleus.
26 Medium
26 of 351
Application +4 | -1
Half life of a radioactive element is 10\mathrm{10} days. The time during which quantity remains 1/10\mathrm{1/10} of initial mass will be
View answer and explanation
Correct option: C
33\mathrm{33} days
Option C (33 days) is the closest to the correct answer because after approximately 3.3 half-lives, the quantity of the radioactive element will be reduced to 1/10 of its initial mass.
27 Medium
27 of 351
Understanding +4 | -1
A nuclear decay is expressed as C611→B511+β++XC_6^{11} \to B_5^{11} + {\beta ^ + } + X. Then the unknown particle X is
View answer and explanation
Correct option: D
Neutrino
Option D (Neutrino) is correct because in positive beta decay, a neutrino is emitted to conserve lepton number.
28 Medium
28 of 351
Application +4 | -1
During mean life of a radioactive element, the fraction that disintegrates is
View answer and explanation
Correct option: C
e−1e\frac{{e - 1}}{e}
Option C (e−1e\frac{{e - 1}}{e}) is correct because it represents the fraction of radioactive material that has decayed after one mean lifetime.
29 Medium
29 of 351
Analyzing +4 | -1
The half-life period of radioactive element xx is same as the mean life time of another radioactive element yy Initially they have the same number of atoms. Then,
View answer and explanation
Correct option: B
y will decay faster than x
Option B is correct because the decay rate is proportional to the decay constant, and since the half-life of X is equal to the mean life of Y, Y has a larger decay constant and thus decays faster.
30 Medium
30 of 351
Application +4 | -1
In the decay 64Cu→64Ni+e++v{ }^{64} {Cu} \rightarrow{ }^{64} {Ni}+e^{+}+{v}, the maximum kinetic energy carried by the positron is found to be 0.650 MeV0.650 \mathrm{~MeV}.

(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?0.150 \mathrm{~MeV}?
(b) What is the momentum of this neutrino in kg m s−1?\mathrm{kg} \mathrm{~m} \mathrm{~s}^{-1}?

Use the formula applicable to a photon.
View answer and explanation
Correct option: A
(a) 500 keV;500 \mathrm{~keV}; (b) 2.67×10−22 kg m s−12.67 \times 10^{-22} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
Option A ((a) 500 keV;500 \mathrm{~keV}; (b) 2.67×10−22 kg m s−12.67 \times 10^{-22} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}) is correct because the neutrino energy is the difference between the maximum positron energy and the given positron kinetic energy, and the momentum is calculated using the photon-like relationship.
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