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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 91-120 of 380 questions
91 Medium
91 of 380
Remembering +4 | -1
Order of nuclear density is
View answer and explanation
Correct option: C
1017 kg/m310^{17} \mathrm{~kg} / \mathrm{m}^3
Option C (1017 kg/m310^{17} \mathrm{~kg} / \mathrm{m}^3) is correct because the density of nuclear matter is extremely high, on the order of 1017 kg/m310^{17} \mathrm{~kg} / \mathrm{m}^3.
92 Medium
92 of 380
Remembering +4 | -1
What is used as a moderator in a nuclear reactor?
View answer and explanation
Correct option: B
Graphite
Option B (Graphite) is correct because graphite is a common moderator used in nuclear reactors to slow down neutrons, increasing the probability of nuclear fission.
93 Easy
93 of 380
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 energy equivalent of a proton's mass calculated using E=mc2E=mc^2.
94 Medium
94 of 380
Application +4 | -1
A radioactive element decays by β\beta-emission. A detector records n\mathrm{n} beta particles in 2 s2 \mathrm{~s} and in next 2 s2 \mathrm{~s} it records 0.75n0.75 \mathrm{n} beta particle. Find mean life correct to nearest whole number.
Given ln⁡2=0.6931\ln 2=0.6931, (n 3=1.09863=1.0986
View answer and explanation
Correct option: B
7 s7 \mathrm{~s}
Option B (7 s) is correct because the mean life is calculated using the decay constant derived from the given beta particle counts over two consecutive time intervals.
95 Medium
95 of 380
Remembering +4 | -1
The rest energy of an electron is
View answer and explanation
Correct option: A
510  KeV510\,\,KeV
Option A (510  KeV510\,\,KeV) is correct because it is the accepted value for the rest energy of an electron calculated using E=mc2E=mc^2.
96 Medium
96 of 380
Application +4 | -1
The decay constant of 238U{ }^{238} \mathrm{U} is 4.9×10−18 s−14 .9 \times 10^{-18} \mathrm{~s}^{-1}.
(a) What is the average-life of 238U?{ }^{238} \mathrm{U}?
(b) What is the half-life of 238U?{ }^{238} \mathrm{U}?
(c) By what factor does the activity of a 238U{ }^{238} \mathrm{U} sample decrease in 9×1099 \times 10^9 years?
View answer and explanation
Correct option: A
(a) 6.49×109 y;6.49 \times 10^9 \mathrm{~y}; (b) 4.5×109 y;4.5 \times 10^9 \mathrm{~y}; (c) 4
Option A ((a) 6.49×109 y;6.49 \times 10^9 \mathrm{~y}; (b) 4.5×109 y;4.5 \times 10^9 \mathrm{~y}; (c) 4) is correct because it correctly calculates the average life, half-life, and activity decrease factor using the given decay constant.
97 Medium
97 of 380
Understanding +4 | -1
The activity of a sample of radioactive material depends on
View answer and explanation
Correct option: C
number of atoms present
Option C is correct because the activity of a radioactive sample is directly proportional to the number of radioactive atoms present.
98 Medium
98 of 380
Application +4 | -1
The half-life of 226Ra{ }^{226} {Ra} is 1602 y1602 \mathrm{~y}. Calculate the activity of 0.1 g0.1 \mathrm{~g} of RaCl2{RaCl}_2 in which all the radium is in the form of 226Ra{ }^{226} {Ra}. Taken atomic weight of Ra{Ra} to be 226 g mol−1226 \mathrm{~g} \mathrm{~mol}^{-1} and that of Cl{Cl} to be 35.5 g mol−135.5 \mathrm{~g} \mathrm{~mol}^{-1}.
View answer and explanation
Correct option: A
2.8×1092.8 \times 10^9 disintegrations s−1\mathrm{s}^{-1}
Option A (2.8×1092.8 \times 10^9 disintegrations s−1\mathrm{s}^{-1}) is correct because it is calculated by finding the number of Radium atoms and multiplying by the decay constant.
99 Medium
99 of 380
Remembering +4 | -1
If λ\lambda is decay constant and NN the number of radioactive nuclei of an element , then the decay rate (RR) of that element is
View answer and explanation
Correct option: B
λN\lambda N
The correct answer is B (λN\lambda N) because the decay rate (R) is directly proportional to the number of radioactive nuclei (N) and the decay constant (λ\lambda).
100 Medium
100 of 380
Application +4 | -1
Calculate the time interval between 33% decay and 67% decay if half -life of a substance is 20 min.
View answer and explanation
Correct option: B
20 min
Option B (20 min) is correct because the time interval between any two percentage decays is equal to the half-life if the percentages are symmetrically placed around 50%.
101 Medium
101 of 380
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.
102 Medium
102 of 380
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.
103 Medium
103 of 380
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.
104 Medium
104 of 380
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.
105 Medium
105 of 380
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.
106 Medium
106 of 380
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.
107 Medium
107 of 380
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.
108 Medium
108 of 380
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.
109 Medium
109 of 380
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.
110 Medium
110 of 380
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.
111 Medium
111 of 380
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.
112 Medium
112 of 380
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.
113 Medium
113 of 380
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.
114 Medium
114 of 380
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.
115 Medium
115 of 380
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.
116 Medium
116 of 380
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.
117 Medium
117 of 380
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.
118 Medium
118 of 380
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.
119 Medium
119 of 380
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.
120 Medium
120 of 380
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.
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