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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 91-120 of 351 questions

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91 Medium
91 of 351
Remembering +4 | -1
In helium nucleus, there are
View answer and explanation
Correct option: C
2  protons\mathrm{2}\,\,\mathrm{protons} and 2  neutrons\mathrm{2}\,\,\mathrm{neutrons}
Option C (2 protons\mathrm{2}\,\mathrm{protons} and 2 neutrons\mathrm{2}\,\mathrm{neutrons}) is correct because a helium nucleus consists of 2 protons and 2 neutrons.
92 Medium
92 of 351
Understanding +4 | -1
Cadmium rods are used in a nuclear reactor for
View answer and explanation
Correct option: C
Absorbing neutrons
Option C is correct because cadmium rods are used to absorb neutrons, controlling the chain reaction in a nuclear reactor.
93 Medium
93 of 351
Application +4 | -1
A sample of radioactive material decays simultaneously by two processes AA and BB with half-lives 12\frac{1}{2} and 14\frac{1}{4} hours respectively. For first half-hour, it decays with process A,A, next one hour with process BB and for a further half an hour with both AA and B.B. If originally there were N0{N}_0 nuclei, find the number of nuclei after 2 hours of such decay
View answer and explanation
Correct option: A
N0(2)8\frac{N_0}{(2)^8}
Option A (N0(2)8\frac{N_0}{(2)^8}) is correct because it accounts for the decay during each time interval with the respective half-lives.
94 Medium
94 of 351
Application +4 | -1
The rest mass of an electron as well as that of positron is 0.51  MeV0.51\,\,MeV. When an electron and positron are annihilate, they produce gamma-rays of wavelength(s)
View answer and explanation
Correct option: A
0.012 A∘0.012\,\mathop \mathrm{A}\limits^ \circ
Option A (0.012 A∘ 0.012\,\mathop \mathrm{A}\limits^ \circ \,) is correct because the wavelength of the gamma-rays produced is calculated using the total energy released during electron-positron annihilation.
95 Medium
95 of 351
Understanding +4 | -1
A nucleus ZAX{ }_Z^A X has mass represented by m(A,Z)m(A, Z), If mpm_p and mnm_n denote the mass of proton and neutron respectively and BE the binding energy (in MeV{MeV} ) then.
View answer and explanation
Correct option: B
BE=[Zmp+(A−Z)mn−m(A,Z)]c2B E=\left[Z m_p+(A-Z) m_n-m(A, Z)\right] c^2
Option B (BE=[Zmp+(A−Z)mn−m(A,Z)]c2B E=\left[Z m_p+(A-Z) m_n-m(A, Z)\right] c^2) is correct because the binding energy is the difference between the mass of the individual nucleons and the mass of the nucleus, multiplied by c2c^2.
96 Medium
96 of 351
Understanding +4 | -1
In radioactive decay process, the negatively charged emitted β−\mathrm{\beta } - particles are
View answer and explanation
Correct option: B
The electrons produced as a result of the decay of neutrons inside the nucleus
Option B (The electrons produced as a result of the decay of neutrons inside the nucleus) is correct because beta decay involves the transformation of a neutron into a proton, an electron, and an antineutrino within the nucleus.
97 Medium
97 of 351
Application +4 | -1
The nuclei of two radioactive isotopes of same substance A236\mathrm{A}^{236} and A234\mathrm{A}^{234} are present in the ratio 4:14: 1 in an ore obtained from Mars. Their half lives are 30 min30 \mathrm{~min} and 60 min60 \mathrm{~min} respectively. Both isotopes are alpha emitters and the activity of the isotope with half life 30 min\mathrm{min} is one Rutherford. Calculate after how much time (in min) their activities will become identical.
View answer and explanation
Correct option: A
180
Option A (180) is correct because it is the time when the activities of both isotopes will be equal, calculated using their half-lives and initial ratio.
98 Medium
98 of 351
Understanding +4 | -1
For a nucleus to be stable, the correct relation between neutron number NN and proton number ZZ is
View answer and explanation
Correct option: D
N≥ZN \ge Z
Option D (N≥ZN \ge Z) is correct because for stability, the number of neutrons is greater than or equal to the number of protons, especially for heavier nuclei.
99 Medium
99 of 351
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}
The correct answer is C. In option C, the mass number and atomic number are not conserved, making the reaction impossible.
100 Medium
100 of 351
Analyzing +4 | -1
10 gm of a radioactive element is kept in a container. The element is β\beta-active. Then after one half-life:

(molar mass of the substance is 100 gm, Avogadro's number =6×1023=6 \times 10^{23} per mole)
View answer and explanation
Correct option: C
If all β\beta particles leave the container then the charge of the substance left is 4800 C
Option C (If all β particles leave the container then the charge of the substance left is 4800 C) is correct. Beta decay doesn't significantly change the mass, and the charge left after beta emission can be calculated.
101 Medium
101 of 351
Understanding +4 | -1
The electron emitted in beta radiation originates from
View answer and explanation
Correct option: C
Decay of a neutron in a nucleus
Option C (Decay of a neutron in a nucleus) is correct because beta radiation involves the transformation of a neutron into a proton and an electron within the nucleus.
102 Medium
102 of 351
Remembering +4 | -1
Atomic number of a nucleus is ZZ and atomic mass is MM. The number of neutron is
View answer and explanation
Correct option: A
M−ZM - Z
Option A (M−ZM - Z) is correct because the number of neutrons is the difference between the mass number and the atomic number.
103 Medium
103 of 351
Understanding +4 | -1
MxM_x and MyM_y denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The QQ-value of a β−\beta^{-}decay is Q1Q_1 and that for a β+\beta^{+}decay is Q2Q_2. If mem_e denotes the mass of an electron, then which of the following statement is correct?
View answer and explanation
Correct option: A
Q1=(Mx−My)c2Q_1=\left(M_x-M_y\right) c^2 and Q2=(Mx−My−2me)c2Q_2=\left(M_x-M_y-2 m_e\right) c^2
Option A (Q1=(Mx−My)c2Q_1=\left(M_x-M_y\right) c^2 and Q2=(Mx−My−2me)c2Q_2=\left(M_x-M_y-2 m_e\right) c^2) is correct because it correctly accounts for the mass differences and electron masses in beta decay Q-value calculations.
104 Medium
104 of 351
Understanding +4 | -1
In gamma ray emission from a nucleus
View answer and explanation
Correct option: B
There is no change in the proton number and the neutron number
Option B is correct because gamma ray emission involves the release of energy from the nucleus, without changing the number of protons or neutrons.
105 Medium
105 of 351
Remembering +4 | -1
In which radiation atomic number and mass number are not
affected?
View answer and explanation
Correct option: B
γ−radiation\gamma-\text{radiation}
Option B (γ−radiation\gamma-\text{radiation}) is correct because gamma radiation involves the emission of photons, which do not change the atomic number or mass number of the nucleus.
106 Medium
106 of 351
Application +4 | -1
7N14_7{N^{14}} is bombarded with 2He4_2\mathrm{H}{\mathrm{e}^4}. The resulting nucleus is 8O17_8{\mathrm{O}^{17}} with the emission of
View answer and explanation
Correct option: C
Proton
Option C (Proton) is correct because the nuclear reaction equation balances with the emission of a proton.
107 Medium
107 of 351
Analyzing +4 | -1
In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be 100:1100: 1. The mean lives of the two isotopes are 4×109yr4 \times 10^9 {yr} and 2×109yr2 \times 10^9 \mathrm{yr}, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal proportion, the age of the rock p×1010yrp \times 10^{10} \mathrm{yr}, where the value of pp is ......... (Take, ratio of the atomic weights of the two isotopes is 1.02:11.02: 1.)
View answer and explanation
Correct option: A
1.834
Option A (1.834) is correct because the age of the rock is calculated to be approximately 1.834×10101.834 \times 10^{10} years using the radioactive decay formula and the given ratios.
108 Medium
108 of 351
Application +4 | -1
When 88Ra236_{88}\mathrm{R}{\mathrm{a}^{236}} decays in a series by emission of 3 α3\,\mathrm{\alpha }-particles and one β\mathrm{\beta }-particle, isotope XX formed is
View answer and explanation
Correct option: A
83X224_{83}{X^{224}}
Option A (83X224_{83}{X^{224}}) is correct because after emitting 3 alpha particles and one beta particle, the resulting isotope has an atomic number of 83 and a mass number of 224.
109 Medium
109 of 351
Application +4 | -1
The binding energy of deuteron is 2.2 Mev and that of 2He4{}_2H{e^4} is 28 Mev. If two deuterons are fused to form one 2He4{}_2H{e^4} then energy released is
View answer and explanation
Correct option: C
23.6 Mev
Option C (23.6 Mev) is correct because the energy released is the difference between the binding energy of the Helium nucleus and twice the binding energy of the deuteron.
110 Medium
110 of 351
Analyzing +4 | -1
Given below are two statements: one is labelled as Assertion A\mathbf{A} and the other is labelled as Reason R\mathbf{R}

Assertion A: The nuclear density of nuclides 510 B{ }_5^{10} \mathrm{~B}, 36Li,2656Fe,1020Ne{ }_3^6 \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne} and 83209Bi{ }_{83}^{209} \mathrm{Bi} can be arranged as ρBiN>ρFeNN>ρNeNN>ρBN>ρLiN\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}_{\mathrm{N}}}^{\mathrm{N}}>\rho_{\mathrm{Ne}_{\mathrm{N}}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}

Reason R\mathbf{R} : The radius RR of nucleus is related to its mass number AA as R=R0A1/3R=R_0 A^{1 / 3}, where R0R_0 is a constant.

In the light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: D
AA is false but RR is true
Option D (AA is false but RR is true) is correct because the nuclear density is approximately constant for all nuclei, and the radius is indeed related to the mass number as stated in the reason.
111 Medium
111 of 351
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.
112 Medium
112 of 351
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.
113 Medium
113 of 351
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.
114 Medium
114 of 351
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.
115 Medium
115 of 351
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.
116 Medium
116 of 351
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.
117 Medium
117 of 351
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.
118 Medium
118 of 351
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.
119 Medium
119 of 351
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.
120 Medium
120 of 351
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.
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