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JEE • PHYSICS • NUCLEI

Nuclei practice · Mixed difficulty

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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 121-150 of 380 questions

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121 Medium
121 of 380
Application +4 | -1
Plutonium decays with half-life of 24000 yr\mathrm{24000 }\,\mathrm{yr}. If plutonium is stored for 7200 yr\mathrm{7200 }\,\mathrm{yr}, the fraction of it that remains is
View answer and explanation
Correct option: A
1/8\mathrm{1/8}
Option A (1/8) is correct because after 3 half-lives, the remaining fraction is (1/2)3.
122 Hard
122 of 380
Remembering +4 | -1
Choose incorrect statement:
View answer and explanation
Correct option: A
Nuclear forces are short ranging and charge
dependent
Option B (γ\gamma-rays) is correct because gamma rays have the shortest wavelength among the given electromagnetic waves.
123 Medium
123 of 380
Understanding +4 | -1
Which can pass through 20  cm20\,\,cm thickness of the steel
View answer and explanation
Correct option: C
γ−\gamma - rays
Option C (γ−\gamma - rays) is correct because gamma rays have the highest penetrating power among the given options.
124 Medium
124 of 380
Analyzing +4 | -1
A radioactive nucleus ZAX_Z^AX undergoes spontaneous decay in the sequence
ZAX→Z−1B→Z−3C→Z−2D,_Z^AX{ \to _{Z - 1}}B{ \to _{Z - 3}}C{ \to _{Z - 2}}D, where ZZ is the atomic number of element XX. The possible decay particles in the sequence are
View answer and explanation
Correct option: A
β+,α,β−{\beta ^ + },\alpha ,{\beta ^ - }
The correct answer is A (β+,α,β−{\beta ^ + },\alpha ,{\beta ^ - }) because a positron decay decreases the atomic number by 1, an alpha decay decreases it by 2, and a beta-minus decay increases it by 1.
125 Medium
125 of 380
Understanding +4 | -1
Assertion: In the process of nuclear fission, the fragments emit two or three neutrons as soon as they are formed and subsequently emit particles.

Reason: As the fragments contain an excess of neutrons over protons, emission of neutrons and particles bring their neutron/proton ratio to stable values.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because nuclear fission releases neutrons to achieve a stable neutron/proton ratio.
126 Medium
126 of 380
Understanding +4 | -1
Statement-I: If a heavy nucleus is split into two medium sized aprts, each of the new nuclei will have more binding energy per nucleon than the original nucleus.

Statement-II: Joining two light nuclei together to give a single nucleus of meidum size means more B.E per nucleon in the new nucleus.
View answer and explanation
Correct option: A
Statement-I and Statement-II are true
Option A is correct because both statements accurately describe the changes in binding energy per nucleon during nuclear fission and fusion.
127 Medium
127 of 380
Understanding +4 | -1
Carbon (Z=6)(Z=6) with mass number 11 decays to boron (Z=5)(Z=5). (a) Is it a β+\beta^{+}-decay or a β−\beta^{-}-decay? (b) The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90%90 \% carbon-11 and 10%10 \% boron-11 (by the number of atoms) converts itself into a mixture of 10%10 \% carbon-11 and 90%90 \% boron- 11?
View answer and explanation
Correct option: A
(a) β+;\beta^{+}; (b) 64 min64 \mathrm{~min}
Option A ((a) β+;\beta^{+}; (b) 64 min64 \mathrm{~min}) is correct because carbon decays to boron via positron emission, and the time for the conversion is approximately 64 minutes.
128 Medium
128 of 380
Application +4 | -1
When a proton, anti-proton annihilate ,the energy released is
View answer and explanation
Correct option: A
1.5  ×10−10 J1.5\,\, \times {10^{ - 10}}\,\mathrm{J}
Option A (1.5×10−10 J1.5 \times {10^{ - 10}}\,\mathrm{J}) is correct because it is the energy released when a proton and anti-proton annihilate, calculated using Einstein's mass-energy equivalence.
129 Medium
129 of 380
Understanding +4 | -1
A radioactive nucleus emits a beta particle. The parent and daughter nuclei are
View answer and explanation
Correct option: D
Isobars
Option D (Isobars) is correct because beta decay changes the atomic number but not the mass number, resulting in isobars.
130 Medium
130 of 380
Application +4 | -1
A radio-isotope has a half-life of 5  years5\,\,years. The fraction of the atoms of this material that would decay in 15  years15\,\,years will be
View answer and explanation
Correct option: C
7/8\mathrm{7/8}
Option C (7/8\mathrm{7/8}) is correct because after 15 years, which is three half-lives, 7/8 of the material will have decayed.
131 Medium
131 of 380
Application +4 | -1
1 mg gold undergoes decay with 2.7 days half-life period,
amount left after 8.1 days is:
View answer and explanation
Correct option: D
0.125 mg
Option D (0.125 mg) is correct because after 8.1 days, which is three half-lives, the remaining amount is (1/2)3 of the initial amount.
132 Medium
132 of 380
Understanding +4 | -1
If a particle is being added to the nucleus of an atom but not changing its chemical properties. Then, these particles are
View answer and explanation
Correct option: B
neutrons
Option B (neutrons) is correct because adding neutrons to the nucleus changes the isotope of the atom but does not alter the number of protons, which determines the chemical properties.
133 Medium
133 of 380
Application +4 | -1
A radioactive β\beta-emission. A detector records nβn \beta-particles in 2s2 s and by next 2s2 s (accumulatively) it records 1.1nβ1.1 n \beta-particle. Number of β\beta-particles recorded by detector after a long time, is
View answer and explanation
Correct option: D
10n9\frac{10 n}{9}
Option D (10n9\frac{{10 n}}{{9}}) is correct because it represents the total number of beta particles recorded after a long time, considering the decay process.
134 Medium
134 of 380
Remembering +4 | -1
In nuclear reactions, we have the conservation of
View answer and explanation
Correct option: D
Mass, energy and momentum
Option D (Mass, energy and momentum) is correct because in nuclear reactions, mass-energy and momentum are always conserved.
135 Medium
135 of 380
Understanding +4 | -1
If M0{M_0} is the mass of an oxygen isotope 8O17_8{\mathrm{O}^{17}} , Mp  and  Mn{M_p}\,\,and\,\,{M_n} are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is
View answer and explanation
Correct option: B
(Mo−8Mp−9Mn)c2\left( {{M_o}-8{M_p} - 9{M_n}} \right){c^2}
The correct answer is B because the nuclear binding energy is the difference between the mass of the individual nucleons and the mass of the nucleus, multiplied by c².
136 Medium
136 of 380
Analyzing +4 | -1
Assertion: In a decay atom number of daughter nucleus reduces by 2 from the parent nucleus.

Reason:An Q-particle carries four units of mass.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is correct explanation of assertion.
Option A is correct. In alpha decay, the atomic number of the daughter nucleus is reduced by 2 from the parent nucleus because an alpha particle carries two protons and two neutrons.
137 Medium
137 of 380
Application +4 | -1
The fossil bone has a 14C:12C^{14}\mathrm{C}{:^{12}}\mathrm{C} ratio, which is [116]\left[ {\frac{1}{{16}}} \right] of that in a living animal bone. If the half-life of 14C^{14}\mathrm{C} is 5730 yr\mathrm{5730 }\,\mathrm{yr}, then the age of the fossil bone is
View answer and explanation
Correct option: C
22920 yr\mathrm{22920 }\,\mathrm{yr}
Option C (22920 yr\mathrm{22920 }\,\mathrm{yr}) is correct because the age of the fossil is determined by calculating the number of half-lives that have passed based on the ratio of 14C^{14}\mathrm{C} to 12C^{12}\mathrm{C}.
138 Medium
138 of 380
Analyzing +4 | -1
Following statements related to radioactivity are given below:
(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.
(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.
(C) Slope of the graph of loge (no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time (τ).\left( \tau \right).
(D) Product of decay constant (λ)\left( \lambda \right) and half-life time (T1/2)\left( {{T_{1/2}}} \right) is not constant.
Choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
(B) and (C) only
Option B ((B) and (C) only) is correct because radioactive decay is independent of physical/chemical conditions, the number of undecayed nuclei decays exponentially, and the slope of the ln(N) vs. time graph is the negative of the decay constant.
139 Medium
139 of 380
Understanding +4 | -1
If Mo\mathrm{M}_{\mathrm{o}} is the mass of isotope 512B,Mp{ }_5^{12} \mathrm{B}, \mathrm{M}_{\mathrm{p}} and Mn\mathrm{M}_{\mathrm{n}} are the masses of proton and neutron, then nuclear binding energy of isotope is:
View answer and explanation
Correct option: A
(5Mp+7Mn−Mo)C2\left(5 \mathrm{M}_{\mathrm{p}}+7 \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{\mathrm{o}}\right) \text{C}^2
Option A is correct because the nuclear binding energy is the difference between the mass of the individual nucleons and the mass of the nucleus, multiplied by C2\text{C}^2.
140 Medium
140 of 380
Application +4 | -1
Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is
View answer and explanation
Correct option: C
5:27\mathrm{5 : 27}
Option C (5 : 27) is correct because it is the correct ratio calculated using the Rydberg formula for the first lines of the Lyman and Balmer series.
141 Medium
141 of 380
Understanding +4 | -1
The correct sequence of the increasing wavelength of the given radiation sources is
View answer and explanation
Correct option: B
Radioactive source, X-ray tube, sodium vapour lamp, crystal oscillator
Option B (Radioactive source, X-ray tube, sodium vapour lamp, crystal oscillator) is correct because it represents the correct order of increasing wavelength.
142 Medium
142 of 380
Application +4 | -1
Two nuclei have their mass numbers in the ratio of 1 : 27. The ratio of their volume is
View answer and explanation
Correct option: B
1:271\mathrm{ }:\mathrm{ }27
Option B (1:271\mathrm{ }:\mathrm{ }27) is correct because the volume of a nucleus is directly proportional to its mass number.
143 Medium
143 of 380
Application +4 | -1
92238U{}_{\mathrm{92}}^{\mathrm{238}}\mathrm{U} emits 8 α−\alpha \mathrm{ - } particle and 6 β−\beta - particles. The neutron/proton ratio in the product nucleus is
View answer and explanation
Correct option: C
62/41
The correct answer is 62/41 because after emission of alpha and beta particles, the neutron/proton ratio is calculated based on the remaining number of neutrons and protons.
144 Medium
144 of 380
Remembering +4 | -1
The subatomic particles proton and neutron fall under the group of
View answer and explanation
Correct option: D
Baryons
Option D (Baryons) is correct because protons and neutrons are classified as baryons.
145 Medium
145 of 380
Application +4 | -1
An artificial radioactive decay series begins with unstable 94241Pu_{94}^{241}Pu. The stable nuclide obtained after eight α−\alpha - decays and five β−\beta - decays is
View answer and explanation
Correct option: A
83209 Bi_{83}^{209}\,Bi
Option A (83209 Bi_{83}^{209}\,Bi) is correct because after 8 alpha decays and 5 beta decays, the resulting nuclide has a mass number of 209 and an atomic number of 83.
146 Medium
146 of 380
Understanding +4 | -1
Fusion process, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. This reasons for this can be traced to the fact
View answer and explanation
Correct option: A
Nuclear forces have short range
Option A (Nuclear forces have short range) is correct because the strong nuclear force, which binds nucleons together, operates only at very short distances, preventing fusion at ordinary temperatures and pressures.
147 Medium
147 of 380
Application +4 | -1
The radius of a spherical nucleus as measured by electron scattering is 3.6 fm. What is the likely mass number of the nucleus?
View answer and explanation
Correct option: A
27
Option A (27) is correct because the mass number is calculated using the formula relating nuclear radius and mass number.
148 Medium
148 of 380
Understanding +4 | -1
For nuclear reaction, select correct statement for released energy
View answer and explanation
Correct option: A
Release energy per mass is more in fusion
Option A (Release energy per mass is more in fusion) is correct because fusion reactions generally release more energy per unit mass than fission reactions.
149 Medium
149 of 380
Application +4 | -1
A nuclear fusion reaction is given by 4(24He)→816O4\left( {{}_2^4He} \right) \to {}_8^{16}O. If atomic mass of He is 4.0026 amu and that of O is 15.9994 amu then the Q-value of the reaction will be approximately
View answer and explanation
Correct option: B
10.2 MeV
Option B (10.2 MeV) is correct because it is the energy released in the nuclear fusion reaction, calculated from the mass defect.
150 Medium
150 of 380
Understanding +4 | -1
Consider the following two statements AA and BB identify the correct answer given
A: Nuclear density is same for all nuclei
B:Radius of the nucleus RR and its mass the number AA are related as A∝R1/6\sqrt A \propto {R^{1/6}}
View answer and explanation
Correct option: C
A\mathrm{A} is true but B\mathrm{B} is false
Statement A is true because nuclear density is approximately constant for all nuclei. Statement B is false because the radius is proportional to A(1/3), not A(1/6).
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