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Nucleus and Radioactivity — NEET Physics PYQ MCQs with Solutions

Free NEET Physics PYQ 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
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: $\beta^+, \alpha, \beta^-$
Q2 — 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: $\frac{A}{A_0} = 2^{-t/t_{1/2}} = 2^{-150/100} = 2^{-3/2} = \frac{1}{2\sqrt{2}}$
Q3 — Nucleus and Radioactivity · easy · 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.
Q4 — 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
Q5 — 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: $R = \lambda N \Rightarrow N = \frac{R \cdot T_{1/2}}{0.693} = \frac{10^{10} \times 2.2 \times 10^9}{0.693} = 3.17 \times 10^{19}$
Q6 — Nucleus and Radioactivity · easy · 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.
Q7 — 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
Q8 — 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: $N_A = N_0e^{-8\lambda t}$, $N_B = N_0e^{-\lambda t}$ $\frac{N_A}{N_B} = e^{-7\lambda t} = \frac{1}{e} \Rightarrow 7\lambda t = 1$ $t = \frac{1}{7\lambda}$
Q9 — Nucleus and Radioactivity · hard · 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
Q10 — Nucleus and Radioactivity · medium · numerical
If radius of the $^{27}_{13}$Al nucleus is taken to be $R_{Al}$, then the radius of $^{125}_{53}$Te nucleus is nearly
A. $\left(\frac{53}{13}\right)^{1/3}R_{Al}$
B. $\frac{5}{3}R_{Al}$  ✓ Correct
C. $\frac{3}{5}R_{Al}$
D. $\left(\frac{13}{53}\right)^{1/3}R_{Al}$
Solution: $R \propto A^{1/3}$ $\frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$
Q11 — Nucleus and Radioactivity · medium · numerical
A radio isotope X with a half life $1.4 \times 10^9$ yr decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1 : 7. The age of the rock is
A. $1.96 \times 10^9$ yr
B. $3.92 \times 10^9$ yr
C. $4.20 \times 10^9$ yr  ✓ Correct
D. $8.40 \times 10^9$ yr
Solution: X : Y = 1 : 7 means $\frac{1}{8}$ of X remains — 3 half-lives. $t = 3 \times 1.4 \times 10^9 = 4.2 \times 10^9$ yr
Q12 — Nucleus and Radioactivity · medium · numerical
The half-life of a radioactive isotope X is 20 yr. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be
A. 40 yr
B. 60 yr  ✓ Correct
C. 80 yr
D. 100 yr
Solution: $\frac{N}{N_0} = \frac{1}{1+7} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$ — 3 half-lives $t = 20 \times 3 = 60$ yr
Q13 — Nucleus and Radioactivity · medium · numerical
A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of $A_1$ and 160 g of $A_2$. The amount of the two in the mixture will become equal after
A. 60 s
B. 80 s
C. 20 s
D. 40 s  ✓ Correct
Solution: In 40 s: $A_1$ (2 half-lives): 40 → 20 → 10 g; $A_2$ (4 half-lives): 160 → 80 → 40 → 20 → 10 g Both equal 10 g after 40 s.
Q14 — Nucleus and Radioactivity · medium · numerical
A radioactive nucleus of mass M emits a photon of frequency $\nu$ and the nucleus recoils. The recoil energy will be
A. $h^2\nu^2/2Mc^2$  ✓ Correct
B. zero
C. $h\nu$
D. $Mc^2 - h\nu$
Solution: Photon momentum $p = \frac{h\nu}{c}$; by momentum conservation the nucleus recoils with the same momentum. Recoil energy $= \frac{p^2}{2M} = \frac{h^2\nu^2}{2Mc^2}$
Q15 — Nucleus and Radioactivity · medium · numerical
The half-life of a radioactive isotope X is 50 yr. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be
A. 200 yr  ✓ Correct
B. 250 yr
C. 100 yr
D. 150 yr
Solution: $\frac{N}{N_0} = \frac{1}{16} = \left(\frac{1}{2}\right)^4$ — 4 half-lives $t = 4 \times 50 = 200$ yr
Q16 — Nucleus and Radioactivity · medium · theory
A nucleus $^{m}_{n}X$ emits one $\alpha$-particle and two $\beta^-$ particles. The resulting nucleus is
A. $^{m-6}_{n}Z$
B. $^{m-4}_{n}X$  ✓ Correct
C. $^{m-4}_{n-2}Y$
D. $^{m-6}_{n-4}Z$
Solution: $\alpha$ emission: A −4, Z −2; two $\beta^-$ emissions: Z +2. Net: mass number m−4, atomic number unchanged — an isotope $^{m-4}_{n}X$.
Q17 — Nucleus and Radioactivity · hard · 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
Q18 — Nucleus and Radioactivity · medium · theory
The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an
A. isobar of parent
B. isomer of parent
C. isotone of parent
D. isotope of parent  ✓ Correct
Solution: Each ($\alpha + 2\beta$) set reduces A by 4 but leaves Z unchanged. Same Z, different A — an isotope of the parent.
Q19 — Nucleus and Radioactivity · medium · theory
In the nuclear decay given below $^{A}_{Z}X \rightarrow\ ^{A}_{Z+1}Y \rightarrow\ ^{A-4}_{Z-1}B^{*} \rightarrow\ ^{A-4}_{Z-1}B$, the particles emitted in the sequence are
A. $\beta, \alpha, \gamma$  ✓ Correct
B. $\gamma, \beta, \alpha$
C. $\beta, \gamma, \alpha$
D. $\alpha, \beta, \gamma$
Solution: X → Y: Z increases by 1 ($\beta^-$); Y → B*: A −4, Z −2 ($\alpha$); B* → B: de-excitation ($\gamma$). Sequence: $\beta, \alpha, \gamma$
Q20 — Nucleus and Radioactivity · medium · numerical
Two radioactive materials $X_1$ and $X_2$ have decay constants $5\lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $\frac{1}{e}$ after a time
A. $\lambda$
B. $\frac{1}{2}\lambda$
C. $\frac{1}{4\lambda}$  ✓ Correct
D. $\frac{e}{\lambda}$
Solution: $\frac{N_1}{N_2} = e^{-4\lambda t} = \frac{1}{e} \Rightarrow 4\lambda t = 1$ $t = \frac{1}{4\lambda}$
Q21 — Nucleus and Radioactivity · medium · numerical
Two radioactive substances A and B have decay constants $5\lambda$ and $\lambda$ respectively. At $t = 0$ they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $\left(\frac{1}{e}\right)^2$ after a time interval
A. $\frac{1}{4\lambda}$
B. $4\lambda$
C. $2\lambda$
D. $\frac{1}{2\lambda}$  ✓ Correct
Solution: $\frac{N_A}{N_B} = e^{-4\lambda t} = e^{-2} \Rightarrow 4\lambda t = 2$ $t = \frac{1}{2\lambda}$
Q22 — Nucleus and Radioactivity · medium · numerical
If the nucleus $^{27}_{13}$Al has a nuclear radius of about 3.6 fm, then $^{125}_{52}$Te would have its radius approximately as
A. 6.0 fm  ✓ Correct
B. 9.6 fm
C. 12.0 fm
D. 4.8 fm
Solution: $R \propto A^{1/3}$ $R_{Te} = 3.6 \times \left(\frac{125}{27}\right)^{1/3} = 3.6 \times \frac{5}{3} = 6$ fm
Q23 — Nucleus and Radioactivity · medium · theory
In radioactive decay process, the negatively charged emitted $\beta$-particles are
A. the electrons present inside the nucleus
B. the electrons produced as a result of the decay of neutrons inside the nucleus  ✓ Correct
C. the electrons produced as a result of collisions between atoms
D. the electrons orbiting around the nucleus
Solution: Electrons do not pre-exist inside the nucleus. In $\beta^-$ decay a neutron transforms into a proton, an electron and an antineutrino — the electron is created at the moment of emission.
Q24 — Nucleus and Radioactivity · medium · numerical
The radius of germanium (Ge) nuclide is measured to be twice the radius of $^{9}_{4}$Be. The number of nucleons in Ge are
A. 73
B. 74
C. 75
D. 72  ✓ Correct
Solution: $R \propto A^{1/3}$: doubling the radius means $A = 9 \times 2^3 = 72$
Q25 — Nucleus and Radioactivity · medium · theory
In a radioactive material the activity at time $t_1$ is $R_1$ and at a later time $t_2$, it is $R_2$. If the decay constant of the material is $\lambda$, then
A. $R_1 = R_2e^{-\lambda(t_1 - t_2)}$  ✓ Correct
B. $R_1 = R_2e^{\lambda(t_1 - t_2)}$
C. $R_1 = R_2\left(\frac{t_2}{t_1}\right)$
D. $R_1 = R_2$
Solution: $R_1 = R_0e^{-\lambda t_1}$ and $R_2 = R_0e^{-\lambda t_2}$ Dividing: $R_1 = R_2e^{-\lambda(t_1 - t_2)}$
Q26 — Nucleus and Radioactivity · medium · numerical
The nuclei of which one of the following pairs of nuclei are isotones?
A. $^{74}_{34}$Se, $^{71}_{31}$Ga  ✓ Correct
B. $^{92}_{42}$Mo, $^{92}_{40}$Zr
C. $^{84}_{38}$Sr, $^{86}_{38}$Sr
D. $^{40}_{20}$Ca, $^{32}_{16}$S
Solution: Isotones have the same number of neutrons (A − Z). $74 - 34 = 40$ and $71 - 31 = 40$ — Se-74 and Ga-71 are isotones.
Q27 — Nucleus and Radioactivity · easy · 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
Q28 — Nucleus and Radioactivity · medium · numerical
A sample of radioactive element has a mass of 10 g at an instant $t = 0$. The approximate mass of this element in the sample after two mean lives is
A. 3.70 g
B. 6.30 g
C. 1.35 g  ✓ Correct
D. 2.50 g
Solution: After $t = 2\tau = \frac{2}{\lambda}$: $M = M_0e^{-\lambda t} = 10e^{-2} = \frac{10}{e^2} \approx 1.35$ g
Q29 — Nucleus and Radioactivity · easy · 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.
Q30 — Nucleus and Radioactivity · easy · 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.