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Nucleus and Radioactivity — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Nucleus and Radioactivity MCQs with step-by-step solutions (66 questions). Part of Nuclei. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Nucleus and Radioactivity · medium · theory
What happens to the mass number and atomic number of an element when it emits $\gamma$-radiation?
A. Mass number decreases by four and atomic number decreases by two
B. Mass number and atomic number remain unchanged ✓ Correct
C. Mass number remains unchanged, while atomic number decreases by one
D. Mass number increases by four and atomic number increases by two
Solution: $\gamma$-radiation is just a high-energy photon — its emission changes neither the atomic number nor the mass number.
Q2 — Nucleus and Radioactivity · medium · theory
$\alpha$-particle consists of
A. 2 electrons, 2 protons and 2 neutrons
B. 2 electrons and 4 protons only
C. 2 protons only
D. 2 protons and 2 neutrons only ✓ Correct
Solution: An $\alpha$-particle is a doubly ionised helium nucleus (He²⁺) — 2 protons and 2 neutrons, with no electrons.
Q3 — Nucleus and Radioactivity · medium · numerical
The half-life of radium is about 1600 yr. Of 100 g of radium existing now, 25 g will remain unchanged after
A. 4800 yr
B. 6400 yr
C. 2400 yr
D. 3200 yr ✓ Correct
Solution: $\frac{25}{100} = \frac{1}{4} = \left(\frac{1}{2}\right)^2$ — 2 half-lives
$t = 2 \times 1600 = 3200$ yr
Q4 — Nucleus and Radioactivity · medium · theory
A nuclear reaction given by $^{A}_{Z}X \rightarrow\ ^{A}_{Z+1}Y +\ ^{0}_{-1}e + \bar{\nu}$ represents
A. fusion
B. fission
C. $\beta$-decay ✓ Correct
D. $\gamma$-decay
Solution: Emission of an electron ($^{0}_{-1}e$) and an antineutrino with Z increasing by 1 is $\beta^-$ decay.
Q5 — Nucleus and Radioactivity · medium · theory
The mass number of a nucleus is
A. sometimes equal to its atomic number ✓ Correct
B. sometimes less than and sometimes more than its atomic number
C. always less than its atomic number
D. always more than its atomic number
Solution: Mass number = protons + neutrons. For ordinary hydrogen (no neutrons) the mass number equals the atomic number; otherwise it is greater. So it is sometimes equal.
Q6 — Nucleus and Radioactivity · medium · numerical
Half-life of a radioactive substance is 12.5 h and its mass is 256 g. After what time, the amount of remaining substance is 1 g?
A. 75 h
B. 100 h ✓ Correct
C. 125 h
D. 150 h
Solution: $\frac{1}{256} = \left(\frac{1}{2}\right)^8$ — 8 half-lives
$t = 8 \times 12.5 = 100$ h
Q7 — Nucleus and Radioactivity · medium · numerical
Half-life period of a radioactive substance is 6 h. After 24 h activity is 0.01 µC, what was the initial activity?
A. 0.04 µC
B. 0.08 µC
C. 0.24 µC
D. 0.16 µC ✓ Correct
Solution: 24 h = 4 half-lives
$R_0 = 0.01 \times 2^4 = 0.16$ µC
Q8 — Nucleus and Radioactivity · medium · theory
Which of the following is positively charged?
A. $\alpha$-particle ✓ Correct
B. $\beta$-particle
C. $\gamma$-rays
D. X-rays
Solution: X-rays and $\gamma$-rays are electromagnetic waves (no charge); $\beta$-particles are negatively charged electrons. The $\alpha$-particle (helium nucleus) is positively charged.
Q9 — Nucleus and Radioactivity · medium · numerical
The half-life of a radioactive material is 3 h. If the initial amount is 300 g, then after 18 h, it will remain
A. 4.68 g ✓ Correct
B. 46.8 g
C. 9.375 g
D. 93.75 g
Solution: 18 h = 6 half-lives
$N = 300 \times \frac{1}{64} = 4.68$ g
Q10 — Nucleus and Radioactivity · medium · theory
The relationship between disintegration constant ($\lambda$) and half-life (T) will be
A. $\lambda = \frac{\log_{10} 2}{T}$
B. $\lambda = \frac{\log_e 2}{T}$ ✓ Correct
C. $\lambda = \frac{T}{\log_e 2}$
D. $\lambda = \frac{\log_2 e}{T}$
Solution: From $0.5N_0 = N_0e^{-\lambda T}$: $\lambda T = \log_e 2$
$\lambda = \frac{\log_e 2}{T}$
Q11 — Nucleus and Radioactivity · medium · theory
Alpha particles are
A. 2 free protons
B. helium atoms
C. singly ionised helium atoms
D. doubly ionised helium atoms ✓ Correct
Solution: An $\alpha$-particle is a helium nucleus — a helium atom stripped of both its electrons, i.e. a doubly ionised helium atom.
Q12 — Nucleus and Radioactivity · medium · theory
A free neutron decays into a proton, an electron and
A. a beta particle
B. an alpha particle
C. an antineutrino ✓ Correct
D. a neutrino
Solution: $^{1}_{0}n \rightarrow\ ^{1}_{1}H + \beta^- + \bar{\nu}$
A free neutron decays into a proton, an electron and an antineutrino.
Q13 — Nucleus and Radioactivity · medium · theory
The most penetrating radiation out of the following is
A. $\gamma$-rays ✓ Correct
B. $\alpha$-particles
C. $\beta$-rays
D. X-rays
Solution: Penetrating power increases with photon energy ($\propto \frac{1}{\lambda}$). $\gamma$-rays have the shortest wavelength, hence the maximum penetrating power.
Q14 — Nucleus and Radioactivity · medium · numerical
The mass number of He is 4 and that for sulphur is 32. The radius of sulphur nuclei is larger than that of helium by
A. $\sqrt{8}$
B. 4
C. 2 ✓ Correct
D. 8
Solution: $R \propto A^{1/3}$
$\frac{R_S}{R_{He}} = \left(\frac{32}{4}\right)^{1/3} = 2$
Q15 — Nucleus and Radioactivity · medium · theory
If the nuclear force between two protons, two neutrons and between proton and neutron is denoted by $F_{pp}$, $F_{nn}$ and $F_{pn}$ respectively, then
A. $F_{pp} \approx F_{nn} \approx F_{pn}$
B. $F_{pp} \neq F_{nn}$ and $F_{pp} = F_{nn}$
C. $F_{pp} = F_{nn} = F_{pn}$ ✓ Correct
D. $F_{pp} \neq F_{nn} \neq F_{pn}$
Solution: Nuclear forces are charge independent — they act between n–n, p–p and n–p pairs with the same strength.
Q16 — Nucleus and Radioactivity · medium · numerical
In the nucleus of $_{11}$Na²³, the number of protons, neutrons and electrons are
A. 11, 12, 0 ✓ Correct
B. 23, 12, 11
C. 12, 11, 0
D. 23, 11, 12
Solution: Protons $= Z = 11$; neutrons $= A - Z = 23 - 11 = 12$; there are no electrons inside the nucleus.
Q17 — Nucleus and Radioactivity · medium · numerical
The half-life of radium is 1600 yr. The fraction of a sample of radium that would remain after 6400 yr is
A. $\frac{1}{4}$
B. $\frac{1}{2}$
C. $\frac{1}{8}$
D. $\frac{1}{16}$ ✓ Correct
Solution: $n = \frac{6400}{1600} = 4$ half-lives
$\frac{N}{N_0} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$
Q18 — Nucleus and Radioactivity · medium · theory
The constituents of atomic nuclei are believed to be
A. neutrons and protons ✓ Correct
B. protons only
C. electrons and protons
D. electrons, protons and neutrons
Solution: A nucleus of mass number A and atomic number Z contains Z protons and (A − Z) neutrons — nucleons only.
Q19 — Nucleus and Radioactivity · medium · theory
Which of the following statements is true for nuclear forces?
A. They obey the inverse square law of distance
B. They obey the inverse third power law of distance
C. They are short range forces ✓ Correct
D. They are equal in strength to electromagnetic forces
Solution: Nuclear forces are short-range (a few fermi), charge-independent, non-central and are the strongest forces in nature — about 100 times the electrostatic force.
Q20 — Nucleus and Radioactivity · medium · numerical
A radioactive element has half-life period 800 yr. After 6400 yr, what amount will remain?
A. $\frac{1}{2}$
B. $\frac{1}{16}$
C. $\frac{1}{8}$
D. $\frac{1}{256}$ ✓ Correct
Solution: $n = \frac{6400}{800} = 8$ half-lives
$\frac{N}{N_0} = \left(\frac{1}{2}\right)^8 = \frac{1}{256}$
Q21 — Nucleus and Radioactivity · medium · numerical
The nucleus $^{115}_{48}$Cd, after two successive $\beta^-$-decay will give
A. $^{115}_{46}$Pa
B. $^{114}_{49}$In
C. $^{113}_{50}$Sn
D. $^{115}_{50}$Sn ✓ Correct
Solution: Each $\beta^-$ decay keeps A the same and raises Z by 1: $48 \rightarrow 49 \rightarrow 50$, giving $^{115}_{50}$Sn.
Q22 — Nucleus and Radioactivity · medium · numerical
A radioactive sample with a half-life of 1 month has the label: 'Activity = 2 microcurie on 1-8-1991'. What would be its activity two months earlier?
A. 1.0 microcurie
B. 0.5 microcurie
C. 4 microcurie
D. 8 microcurie ✓ Correct
Solution: Two months = 2 half-lives. Going backwards the activity doubles per half-life:
$2 \times 2^2 = 8$ microcurie
Q23 — Nucleus and Radioactivity · medium · numerical
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is
A. 15
B. 30
C. 45
D. 60 ✓ Correct
Solution: After 40% decay: $N_1 = 0.6N_0$; after 85% decay: $N_2 = 0.15N_0$
$\frac{N_2}{N_1} = \frac{0.15}{0.6} = \frac{1}{4} = \left(\frac{1}{2}\right)^2$ — two half-lives
$t = 2 \times 30 = 60$ min
Q24 — Nucleus and Radioactivity · medium · numerical
The activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5$ min. The time (in minute) at which the activity reduces to half its value is
A. $\log_e 2/5$
B. $\frac{5}{\log_e 2}$
C. $5\log_{10} 2$
D. $5\log_e 2$ ✓ Correct
Solution: Activity drops to $\frac{1}{e}$ in 5 min, so the mean life $\tau = \frac{1}{\lambda} = 5$ min.
Half-life $= \tau\log_e 2 = 5\log_e 2$ min
Q25 — Nucleus and Radioactivity · medium · theory
A radioactive nucleus $^{A}_{Z}X$ undergoes spontaneous decay in the sequence $^{A}_{Z}X \rightarrow\ _{Z-1}B \rightarrow\ _{Z-3}C \rightarrow\ _{Z-2}D$, where Z is the atomic number of element X. The possible decay particles in the sequence are
A. $\alpha, \beta^-, \beta^+$
B. $\alpha, \beta^+, \beta^-$
C. $\beta^+, \alpha, \beta^-$ ✓ Correct
D. $\beta^-, \alpha, \beta^+$
Solution: $\beta^+$ decay lowers Z by 1 (Z → Z−1); $\alpha$ decay lowers Z by 2 (Z−1 → Z−3); $\beta^-$ decay raises Z by 1 (Z−3 → Z−2).
Sequence: $\beta^+, \alpha, \beta^-$
Q26 — Nucleus and Radioactivity · medium · numerical
The half-life of a radioactive nuclide is 100 h. The fraction of original activity that will remain after 150 h would be
A. $\frac{1}{2}$
B. $\frac{1}{2\sqrt{2}}$ ✓ Correct
C. $\frac{2}{3}$
D. $\frac{2}{3\sqrt{2}}$
Solution: $\frac{A}{A_0} = 2^{-t/t_{1/2}} = 2^{-150/100} = 2^{-3/2} = \frac{1}{2\sqrt{2}}$
Q27 — Nucleus and Radioactivity · medium · numerical
The half-life of a radioactive sample undergoing $\alpha$-decay is $1.4 \times 10^{17}$ s. If the number of nuclei in the sample is $2.0 \times 10^{21}$, the activity of the sample is nearly
A. $10^4$ Bq ✓ Correct
B. $10^5$ Bq
C. $10^6$ Bq
D. $10^3$ Bq
Solution: Activity $= \lambda N = \frac{0.693}{T_{1/2}}N = \frac{0.693 \times 2 \times 10^{21}}{1.4 \times 10^{17}} \approx 10^4$ Bq
Q28 — Nucleus and Radioactivity · medium · numerical
The rate of radioactive disintegration at an instant for a radioactive sample of half life $2.2 \times 10^9$ s is $10^{10}$ s⁻¹. The number of radioactive atoms in that sample at that instant is
A. $3.17 \times 10^{20}$
B. $3.17 \times 10^{17}$
C. $3.17 \times 10^{18}$
D. $3.17 \times 10^{19}$ ✓ Correct
Solution: $R = \lambda N \Rightarrow N = \frac{R \cdot T_{1/2}}{0.693} = \frac{10^{10} \times 2.2 \times 10^9}{0.693} = 3.17 \times 10^{19}$
Q29 — Nucleus and Radioactivity · medium · numerical
For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
A. 30
B. 10
C. 20 ✓ Correct
D. 15
Solution: Nuclei left undecayed $= 600 - 450 = 150 = \frac{600}{4}$ — that is 2 half-lives.
$t = 2 \times 10 = 20$ min
Q30 — Nucleus and Radioactivity · medium · numerical
Radioactive material A has decay constant $8\lambda$ and material B has decay constant $\lambda$. Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{e}$?
A. $\frac{1}{\lambda}$
B. $\frac{1}{7\lambda}$ ✓ Correct
C. $\frac{1}{8\lambda}$
D. $\frac{1}{9\lambda}$
Solution: $N_A = N_0e^{-8\lambda t}$, $N_B = N_0e^{-\lambda t}$
$\frac{N_A}{N_B} = e^{-7\lambda t} = \frac{1}{e} \Rightarrow 7\lambda t = 1$
$t = \frac{1}{7\lambda}$