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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 211-240 of 380 questions

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211 Easy
211 of 380
Understanding +4 | -1
Statement-I: Positive values of packing fraction implies a large value of binding energy.

Statement-II : The difference between the mass of the nucleus and the mass number of the nucleus is called packing fraction.
View answer and explanation
Correct option: C
Statement-I is true and Statement-II is false
Option D (Statement-I is false and Statement-II is true) is correct because a positive packing fraction implies a smaller binding energy, and the packing fraction is related to the mass defect, not the mass number.
212 Medium
212 of 380
Application +4 | -1
What is approximately the ratio of nuclear radii of gold isotope 79238Au_{79}^{238}Au and silver isotope 47108Ag_{47}^{108}Ag.
View answer and explanation
Correct option: A
1.23
Option A (1.23) is correct because it is the approximate ratio calculated using the formula relating nuclear radius and mass number.
213 Medium
213 of 380
Understanding +4 | -1
Choose the WRONG statement.
View answer and explanation
Correct option: D
The nuclear force becomes very strong if
the nucleus contains a large number of
nucleons
Option D (The nuclear force becomes very strong if the nucleus contains a large number of nucleons) is the WRONG statement because while nuclear force is strong, adding too many nucleons can lead to instability.
214 Medium
214 of 380
Understanding +4 | -1
Mass of the nucleons together in a heavy nucleus is
View answer and explanation
Correct option: A
Greater than mass of nucleus
Option A is correct because the mass of the nucleons (protons and neutrons) when separate is greater than the mass of the nucleus they form due to the binding energy.
215 Medium
215 of 380
Understanding +4 | -1
Two nuclei hare their mass numbers in the ratio of 1 : 3, the ratio of their nuclear densities would be
View answer and explanation
Correct option: B
0.042361111
Option B is correct because nuclear density is independent of mass number.
216 Medium
216 of 380
Understanding +4 | -1
The decay of a proton to neutron is
View answer and explanation
Correct option: B
possible only inside the nucleus
Option B (possible only inside the nucleus) is correct because the neutron is more massive than the proton, and this mass difference can be compensated by the binding energy inside the nucleus.
217 Medium
217 of 380
Understanding +4 | -1
When the number of nucleons in nuclei increase, the binding energy per nucleon
View answer and explanation
Correct option: D
First increases and then decreases with increases of mass number
Option D (First increases and then decreases with increases of mass number) is correct because the binding energy per nucleon initially increases, reaches a peak, and then gradually decreases as the mass number increases.
218 Medium
218 of 380
Understanding +4 | -1
Mass numbers of two nuclei are in the ratio of 4:34: 3. Their nuclear densities will be in the ratio of
View answer and explanation
Correct option: C
1:11: 1
Option C is correct because nuclear density is independent of the mass number.
219 Medium
219 of 380
Application +4 | -1
A certain substance decays 132\frac{1}{32} of its initial activity in 50 days. Calculate its half-life.
View answer and explanation
Correct option: C
10
Option C (10) is correct because the half-life is calculated by finding the number of half-lives in the given time and dividing the total time by that number.
220 Medium
220 of 380
Application +4 | -1
Consider a radioactive substance which decays, following the law of disintegration. At an instant t1t_1 its activity is RR and at time t2t_2 it becomes 1R\frac{1}{R}. Find the average life of radioactive substance, if t2>t1t_2 > t_1.
View answer and explanation
Correct option: C
(t2−t1)2ln⁡(R)\frac{\left(t_2-t_1\right)}{2 \ln (R)}
Option C ((t2−t1)2ln⁡(R)\frac{\left(t_2-t_1\right)}{2 \ln (R)}) is correct because it correctly relates the average life to the given activities at times t1t_1 and t2t_2.
221 Medium
221 of 380
Application +4 | -1
10  g10\,\,g of radioactive material of half-life 15 year\mathrm{15 }\,\mathrm{year} is kept in store for 20 years\mathrm{20 }\,\mathrm{years}. The disintegrated material is
View answer and explanation
Correct option: C
6.03  g6.03\,\,g
Option C (6.03 g6.03\,g) is correct because after 20 years, approximately 6.03g of the radioactive material will have disintegrated.
222 Medium
222 of 380
Application +4 | -1
The half-life of radioactive Radon is 3.8  days\mathrm{3}.\mathrm{8}\,\,\mathrm{days}. The time at the end of which (1/20\mathrm{1/20})th of the Radon sample will remain undecayed is (given log⁡10 e=0.4343{\log _{10}}\,e = 0.4343)
View answer and explanation
Correct option: B
16.5 days\mathrm{16}\mathrm{.5 }\,\mathrm{days}
Option B (16.5 days\mathrm{16}\mathrm{.5 }\,\mathrm{days}) is correct because it is calculated using the half-life formula and the given logarithmic value.
223 Medium
223 of 380
Understanding +4 | -1
Which of the following cannot be emitted by radioactive substances during their decay?
View answer and explanation
Correct option: A
Protons
Option A (Protons) is correct because protons are not typically emitted during radioactive decay processes such as alpha, beta, or gamma decay.
224 Medium
224 of 380
Understanding +4 | -1
The ratio of the density of oxygen nucleus (816O)\left({ }_{8}^{16} \mathrm{O}\right) and helium nucleus (24He)\left({ }_{2}^{4} \mathrm{He}\right) is
View answer and explanation
Correct option: C
1:11: 1
Option C (1:1) is correct because nuclear density is approximately constant for all nuclei, independent of mass number.
225 Medium
225 of 380
Application +4 | -1
Half-life of a radioactive substance is 20  minutes20\,\,minutes. Difference between points of time when it is 33  %\mathrm{33}\,\,\mathrm{\% } disintegrated and 67  %\mathrm{67}\,\,\mathrm{\% } disintegrated is approximately
View answer and explanation
Correct option: B
20  min⁡20\,\,\min
Option B (20  min⁡20\,\,\min) is correct because it is the approximate difference in time between 33% and 67% disintegration, calculated using the half-life formula.
226 Medium
226 of 380
Understanding +4 | -1
Nuclear fission can be explained based on
View answer and explanation
Correct option: B
Liquid drop model
Option B (Liquid drop model) is correct because the liquid drop model of the nucleus explains nuclear fission by treating the nucleus as a liquid drop.
227 Easy
227 of 380
Understanding +4 | -1
Choose the best option for binding energy/nucleon if nuclear energy is released in fission. Binding energy is
View answer and explanation
Correct option: C
larger for fission fragments than for parent nucleus
Option C (larger for fission fragments than for parent nucleus) is correct because nuclear energy is released in fission due to the higher binding energy per nucleon in the fragments.
228 Medium
228 of 380
Understanding +4 | -1
The rad is the correct unit used to report the measurement of
View answer and explanation
Correct option: C
The biological effect of radiation
Option C (The biological effect of radiation) is correct because the rad is a unit used to measure the absorbed dose of radiation, which is related to its biological effect.
229 Medium
229 of 380
Understanding +4 | -1
The mass of an α\alpha-particle is
View answer and explanation
Correct option: A
Less than the sum of masses of two protons and two neutrons
Option D (Equal to sum of masses of two protons and two neutrons) is incorrect because the mass of an alpha particle is less than the sum of the masses of its constituent protons and neutrons due to the binding energy.
230 Medium
230 of 380
Application +4 | -1
Calculate binding energy of helium nuclei

Given mp=1.0073um_p=1.0073 \mathrm{u}

mn=1.0087u\mathrm{m}_{\mathrm{n}}=1.0087 \mathrm{u}

Nuclear mass of He4=4.0015u\mathrm{He}^4=4.0015 \mathrm{u}
View answer and explanation
Correct option: A
28.4MeV28.4 M e V
Option A (28.4MeV28.4 M e V) is correct because it is the binding energy calculated using the mass defect and the energy equivalent.
231 Medium
231 of 380
Application +4 | -1
The Binding energy per nucleon of 37Li{ }_3^7 \mathrm{Li} and 24He{ }_2^4 \mathrm{He} nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction 37Li+11H→24He+24He+Q,{ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}, the value of energy 'Q' released is:
View answer and explanation
Correct option: D
17.3 MeV
Option D (17.3 MeV) is correct because it is the energy released in the nuclear reaction, calculated from the difference in binding energies.
232 Medium
232 of 380
Understanding +4 | -1
If in a nuclear fusion process, the masses of the fusing nuclei be m1{m_1} and m1{m_1} and the mass of the resultant nucleus be m3{m_3}, then
View answer and explanation
Correct option: C
m3<  (m1+  m2){m_3} < \,\,\left( {{m_1} + \,\,{m_2}} \right)
Option C is correct because in nuclear fusion, some mass is converted into energy, resulting in the mass of the resultant nucleus being less than the sum of the masses of the fusing nuclei.
233 Medium
233 of 380
Application +4 | -1
The decay rate of sample of radioactive nuclei I128{I^{128}} after time 4 min and 36 min is 322 and 161 counts/second respectively. The half-life for this radioactive nuclide is
View answer and explanation
Correct option: B
32 min
The correct answer is 32 min. The half-life is calculated by finding the time it takes for the decay rate to halve.
234 Medium
234 of 380
Understanding +4 | -1
A β−\beta^{-}particle is emitted by a radioactive nucleus at the time of conversion of a :-
View answer and explanation
Correct option: A
Neutron into a proton
Option A (Neutron into a proton) is correct because a β−\beta^{-} particle emission occurs when a neutron in the nucleus transforms into a proton, releasing an electron (β−\beta^{-} particle) and an antineutrino.
235 Medium
235 of 380
Remembering +4 | -1
The element with maximum value of binding
energy per nucleon is
View answer and explanation
Correct option: A
iron
Option A (iron) is correct because iron (Fe) has the highest binding energy per nucleon among common elements, making it the most stable nucleus.
236 Medium
236 of 380
Analyzing +4 | -1
Assertion: Density of all the nuclei is same.

Reason: Radius of nucleus is directly proportional to the cube root ofmass number.
View answer and explanation
Correct option: B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
Option B is correct. Both the assertion and the reason are true, but the reason is not the correct explanation of the assertion. The constant nuclear density arises from the short-range nature of the strong nuclear force, not directly from the radius formula.
237 Medium
237 of 380
Analyzing +4 | -1
If the end A\mathrm{A} of a wire is irradiated with α\alpha-rays and the other end BB is irradiated with β\beta-rays. Then
View answer and explanation
Correct option: A
a current will flow from AA to BB
A current will flow from A to B because alpha particles are positively charged and beta particles are negatively charged, creating a potential difference.
238 Medium
238 of 380
Remembering +4 | -1
Most suitable element for nuclear fission is the element with atomic number near
View answer and explanation
Correct option: C
52\mathrm{52}
Option D is incorrect. Most suitable element for nuclear fission is the element with atomic number near 92.
239 Medium
239 of 380
Understanding +4 | -1
Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half-life of 1yr1 {yr}. After 1yr1 {yr},
View answer and explanation
Correct option: C
The containers will in general have different numbers of the the atoms of the material but their average will be close to 5000
Option C is correct because radioactive decay is a random process, so the number of atoms remaining in each container will vary, but the average will be close to 5000.
240 Medium
240 of 380
Remembering +4 | -1
The decay constant of a radioactive sample is λ\lambda.The half-life and mean life of the sample are respectively given by
View answer and explanation
Correct option: B
log⁡e2λ  and  1λ\frac{{{{\log }_e}2}}{\lambda }\,\,\mathrm{and}\,\,\frac{1}{\lambda }
Option B is correct because the half-life is given by log⁡e2λ\frac{{\log_e 2}}{\lambda} and the mean life is given by 1λ\frac{1}{\lambda}.
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