Skip to main content
Back to NUCLEI

Your Progress

0 / 210 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • NUCLEI

Nuclei practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 121-150 of 210 questions

Preparing practice and saved progress…

121 Easy
121 of 210
Application +4 | -1
What are the respective number of α\alpha and β−\beta^- particles emitted in the following radioactive decay?
90200X→ 82172Y_{90}^{200}X \to \,_{82}^{172}Y
View answer and explanation
Correct option: A
7 and 6
Option A (7 and 6) is correct because the emission of an alpha particle reduces the mass number by 4 and the atomic number by 2, while the emission of a beta-minus particle increases the atomic number by 1 without changing the mass number.
122 Medium
122 of 210
Remembering +4 | -1
Hydrogen bomb is based on which of the following phenomenon
View answer and explanation
Correct option: B
Nuclear fusion
Option B is correct because a hydrogen bomb releases energy through uncontrolled nuclear fusion, primarily the fusion of isotopes of hydrogen (deuterium and tritium) into helium.
123 Medium
123 of 210
Analyzing +4 | -1
In the nuclear decay given below:

ZAX⟶Z+1AY⟶Z−1A−4 B∗⟶Z−1A−4 B, { }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{\mathrm{Z}+1}^{\mathrm{A}}\mathrm{Y} \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{~B}^* \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{~B} \text {, }

the particles emitted in the sequence are
View answer and explanation
Correct option: D
β,α,γ\beta, \alpha, \gamma
Option D (β, α, γ) is correct because the first decay increases the atomic number by 1 (beta decay), the second decay decreases the mass number by 4 and atomic number by 2 (alpha decay), and the third decay releases energy without changing A or Z (gamma decay).
124 Medium
124 of 210
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 equivalence and momentum are always conserved.
125 Medium
125 of 210
Remembering +4 | -1
Select the true statement from the following. Nuclear force is
View answer and explanation
Correct option: A
Strong, short range and charge independent force.
Option A (Strong, short range and charge independent force) is correct because nuclear forces are strong, act over short distances, and are independent of the charge of the nucleons.
126 Medium
126 of 210
Remembering +4 | -1
Which is the correct expression for half-life
View answer and explanation
Correct option: D
(t)1/2=2.303  log⁡2λ{\left( t \right)_{1/2}} = \frac{{2.303\,\,\log 2}}{\lambda }
Option D ((t)1/2=2.303  log⁡2λ{\left( t \right)_{1/2}} = \frac{{2.303\,\,\log 2}}{\lambda }) is correct because it represents the correct formula for half-life, derived from the radioactive decay law.
127 Medium
127 of 210
Remembering +4 | -1
Sun maintains its shining because of the
View answer and explanation
Correct option: C
Fusion of hydrogen nuclei
Option C is correct because the sun's energy is produced by nuclear fusion reactions in its core, where hydrogen nuclei fuse to form helium, releasing tremendous energy.
128 Medium
128 of 210
Application +4 | -1
At t=0t = 0, a radioactive substance has a mass m. Its half-life is 10 minutes. When t=t1t = {t_1}, the amount of substance disintegrated is m/5 and when t=t2t = {t_2}, the amount of substance disintegrated is 3m/5. Then the time interval (t2−t1)({t_2} - \mathrm{ }{t_1}) is
View answer and explanation
Correct option: A
10 minutes
Option A (10 minutes) is correct because the time interval between the disintegration of m/5 and 3m/5 is equal to one half-life.
129 Hard
129 of 210
Application +4 | -1
When a sample of atoms is irradiated by neutrons, radioactive atoms are produced at a constant rate RR, which decay with decay constant λ\lambda. The number of radioactive atoms accumulated after an irradiation time tt is given by
View answer and explanation
Correct option: C
N(t)=Rλ(1−e−λt)N(t)=\frac{R}{\lambda}\left(1-e^{-\lambda t}\right)
Option C (N(t)=Rλ(1−e−λt)N(t)=\frac{R}{\lambda}\left(1-e^{-\lambda t}\right)) is correct because it represents the balance between the production and decay of radioactive atoms over time.
130 Medium
130 of 210
Understanding +4 | -1
Mark the correct statement
View answer and explanation
Correct option: A
Nuclei of different elements can have the same number of neutrons
Option A (Nuclei of different elements can have the same number of neutrons) is correct because isotopes of different elements can indeed have the same number of neutrons.
131 Medium
131 of 210
Remembering +4 | -1
π\pi mesons can be
View answer and explanation
Correct option: D
π+{\pi ^ + }, π−{\pi ^ - } or π0{\pi ^0}
The correct answer is D because pions can have positive, negative, or neutral charges.
132 Easy
132 of 210
Application +4 | -1
If a proton is completely converted into energy, the amount of energy produced will be
View answer and explanation
Correct option: D
931×106 eV931 \times 10^6 \mathrm{~eV}
Option D (931×106 eV931 \times 10^6 \mathrm{~eV}) is correct because it is the approximate energy equivalent of the mass of a proton, calculated using E=mc2 and converting to electron volts.
133 Medium
133 of 210
Understanding +4 | -1
Two nuclei have their mass numbers in the ratio of 1:3\mathrm{1:3}. The ratio of their nuclear densities would be
View answer and explanation
Correct option: D
1:1\mathrm{1:1}
Option D (1:1\mathrm{1:1}) is correct because nuclear density is independent of mass number.
134 Medium
134 of 210
Analyzing +4 | -1
Statement I : Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.

Statement II : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z\mathrm{Z}.
View answer and explanation
Correct option: D
Statement - I is true, statement - II is false
Statement I is true because fission and fusion release energy. Statement II is false because binding energy per nucleon decreases for heavy nuclei with increasing Z.
135 Hard
135 of 210
Application +4 | -1
In a nuclear fission reaction of an isotope of mass MM, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect ΔM\Delta M will be:
View answer and explanation
Correct option: D
c2ΔMMc \sqrt{\frac{2 \Delta M}{M}}
Option D (c2ΔMMc \sqrt{\frac{2 \Delta M}{M}}) is correct because it is derived from the conservation of energy and momentum in the nuclear fission.
136 Medium
136 of 210
Remembering +4 | -1
Which of the following is in the increasing order for penetrating power
View answer and explanation
Correct option: A
α,  β,  γ\alpha ,\,\,\beta ,\,\,\gamma
Option A (α,  β,  γ\alpha ,\,\,\beta ,\,\,\gamma) is correct because alpha particles have the lowest penetrating power, followed by beta particles, and then gamma rays have the highest.
137 Medium
137 of 210
Analyzing +4 | -1
Assertion : Size of nucleus is constant in electron scattering or gamma scattering.

Reason : Electron scattering or gamma scattering is controlled by distribution of charge in
nucleus.
View answer and explanation
Correct option: C
If Assertion is True but the reason is false
Option C (If Assertion is True but the reason is false) is correct because the size of the nucleus remains constant during scattering experiments, but the scattering is controlled by the distribution of charge AND matter in the nucleus, not just charge.
138 Medium
138 of 210
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 positron emission, a neutrino is emitted to conserve lepton number.
139 Medium
139 of 210
Application +4 | -1
If half life of radium is 77 days\mathrm{77 }\,\mathrm{days}. Its decay constant in day will be
View answer and explanation
Correct option: B
9  ×  10−3/day9\,\, \times \,\,{10^{ - 3}}/\mathrm{day}
Option B (9 × 10−3/day9\, \times \,{10^{ - 3}}/\mathrm{day}) is correct because the decay constant is calculated using the half-life formula.
140 Medium
140 of 210
Analyzing +4 | -1
Which of the following nuclear reactions is not possible?
View answer and explanation
Correct option: C
511Be+11H⟶49Be+24He{ }_5^{11} \mathrm{Be}+{ }_1^1 \mathrm{H} \longrightarrow{ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He}
Option C (511Be+11H⟶49Be+24He{ }_5^{11} \mathrm{Be}+{ }_1^1 \mathrm{H} \longrightarrow{ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He}) is correct because it violates the conservation of atomic number (charge).
141 Medium
141 of 210
Understanding +4 | -1
The volume of a nucleus is directly proportional to
(where A=A = mass number of the nucleus)
View answer and explanation
Correct option: A
AA
Option A (AA) is correct because the volume of a nucleus is directly proportional to its mass number (A).
142 Medium
142 of 210
Understanding +4 | -1
Consider the following statements
S1\mathrm{S1} : The nuclear force is independent of the charge of nucleons
S2\mathrm{S2} : The number of nucleons in the nucleus of an atom is equal to the number of electrons in the atom
S3\mathrm{S3} : All nuclei have masses that are less than the sum of the masses of constituent nucleons
S4\mathrm{S4} : Nucleons belong to the family of leptons while electrons are members of the family of hadrons
Choose the correct statement (s\mathrm{s}) from these
View answer and explanation
Correct option: D
S1\mathrm{S1} and S3\mathrm{S3}
S1 is true because the nuclear force is charge-independent. S3 is true because the mass of a nucleus is less than the sum of its constituent nucleons due to the binding energy.
143 Medium
143 of 210
Understanding +4 | -1
Which one of the following nuclear reaction is a source of energy in the sun?
View answer and explanation
Correct option: B
2He3+2He3→2He4+1H1+1H1_2\mathrm{H}{\mathrm{e}^3}{ + _2}\mathrm{H}{\mathrm{e}^3}{ \to _2}\mathrm{H}{\mathrm{e}^4}{ + _1}{\mathrm{H}^1}{ + _1}{\mathrm{H}^1}
Option B is correct because the fusion of Helium-3 nuclei to form Helium-4 and protons is a significant part of the proton-proton chain, which is the primary energy source in the sun.
144 Medium
144 of 210
Analyzing +4 | -1
The particle AA is converted into CC, viavia following reaction.
A→B+2He4A \to B{ + _2}\mathrm{H}{\mathrm{e}^4}
B→C  +  2e−B \to C\,\, + \,\,2{e^ - }
View answer and explanation
Correct option: B
AA and CC are isotopes
Option B (AA and CC are isotopes) is correct because isotopes have the same atomic number but different mass numbers, which is the case after the alpha and beta decays.
145 Medium
145 of 210
Analyzing +4 | -1
Radioactive decay will occur as follows

8622Rn→84216PO+25He half life =55 s84216Po→212 Pb+24He half life =0.66 s8282 Pb→82212BL+λ∘e Half life =10.6 h\begin{aligned} & { }_{86}^{22} {Rn} \rightarrow{ }_{84}^{216} {PO}+{ }_2^5 {He} \text { half life }=55 {~s} \\ & { }_{84}^{216} {Po} \rightarrow{ }^{212} {~Pb}+{ }_2^4 {He} \text { half life }=0.66 {~s} \\ & { }_{82}^{82} {~Pb} \rightarrow{ }_{82}^{212} {BL}+\lambda^{\circ} e \text { Half life }=10.6 {~h} \end{aligned}

If a certain mass of radon (Rn=220)({Rn}=220) is allowed to decay in a certain container; then after 5 minutes the element with the greater mass will be
View answer and explanation
Correct option: C
Lead
Lead is the element with the greatest mass after 5 minutes because it has a much longer half-life than Radon and Polonium, allowing it to accumulate.
146 Easy
146 of 210
Understanding +4 | -1
Which of the following are suitable for the fusion process?
View answer and explanation
Correct option: A
Light nuclei.
Option A is correct because light nuclei require less energy to overcome electrostatic repulsion and fuse together, releasing energy in the process, as seen in the fusion reactions in stars.
147 Medium
147 of 210
Application +4 | -1
If a H2{H_2} nucleus is completely converted into energy, the energy produced will be around
View answer and explanation
Correct option: B
938  MeV938\,\,MeV
Option B (938 MeV938\,MeV) is correct because the energy produced is approximately equal to the mass of a proton, which is around 938 MeV.
148 Medium
148 of 210
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 proportional to its mass number.
149 Medium
149 of 210
Analyzing +4 | -1
It is observed that only 0.39%0.39 \% of the original radioactive sample remains undecayed after eight hours select the incorrect statement :-
View answer and explanation
Correct option: D
If the number of radioactive nuclei of this substance at a given instant is 10810^8 then the no. left after 30 min30 {~min} would be 2×107\sqrt{2} \times 10^7
Option D (If the number of radioactive nuclei of this substance at a given instant is 10810^8 then the no. left after 30 min30 {~min} would be 2×107\sqrt{2} \times 10^7) is incorrect because the half-life is 1 hour, and after 30 minutes (half a half-life), the remaining amount is not simply related by a factor of 2\sqrt{2}.
150 Medium
150 of 210
Analyzing +4 | -1
Consider the binding energy of 17Cl35{ }_{17} {Cl}^{35} and 15P31{ }_{15} {P}^{31} are 287.67 MeV and 262.48MeV262.48 {MeV}, respectively. Then,
View answer and explanation
Correct option: B
15P31{ }_{15} {P}^{31} is more stable than 17Cl35{ }_{17} {Cl}^{35}
Option B is correct because the stability of a nucleus is determined by its binding energy per nucleon, and P-31 has a higher binding energy per nucleon than Cl-35.
Revision notes and key takeaways

Review NUCLEI in your textbook or class notes, then use this set to check your understanding.

Showing 121-150 of 210 questions

Your practice

0 / 210 answered · 210 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.