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Nuclei MCQs for NEET — Physics Questions with Answers

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The half life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially both of them have the same number of atoms. Then

X and Y have the same decay rate initially

X and Y decay at the same rate always

Y will decay at a faster rate than X

X will decay at a faster rate than Y

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Explanation

$\Rightarrow \frac{0.693}{λ_{x}} = \frac{1}{λ_{y}} \Rightarrow λ_{x} = 0.693 λ_{y} or λ_{x} < λ_{y}$

Also rate of decay = $λ$ N
Initially number of atoms (N) of both are equal but since $λ_{y} > λ_{x}$ therefore, y will decay at a faster rate than x.

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After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is

6000

9000

3000

24000

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Explanation

(d) Here the activity of the radioactive sample reduces to half in 140 days. Therefore, the half life of the sample is 140 days. 280 days is it’s two half lives. So before two half lives it’s activity was ( $2^{2} \times$ present activity).
$∴$ Initial activity = $2^{2} \times$ 6000=24000 dps

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A radioactive sample is $α$ -emitter with half life 138.6 days is observed by a student to have 2000 disintegration/sec. The number of radioactive nuclei for given activity are

$3.45 \times 10^{10}$

$1 \times 10^{10}$

$3.45 \times 10^{15}$

$2.75 \times 10^{11}$

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Explanation

(a) Activity of substance that has 2000 disintegration/sec

The number of radioactive nuclei having activity A
$N = \frac{A}{λ} = \frac{2000 \times T_{1 / 2}}{\log_{e} 2} = \frac{2000 \times 138.6 \times 24 \times 3600}{0.693} = 3.45 \times 10^{10}$

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The ratio of radii of nuclei ${}_{13}^{27}{Al}$ and ${}_{52}^{A}X$ is 3 : 5. The number of neutrons in the nuclei of X will be

125

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Explanation

(b) $r \propto A^{1 / 3} \Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$

$\Rightarrow \frac{3}{5} = {(\frac{27}{A})}^{1 / 3} \Rightarrow \frac{27}{125} = \frac{27}{A} \Rightarrow A = 125$

Number of nuclei in atom X = A-52 = 125-52 = 73

NEET 2023

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The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\dfrac{1}{16}\right)^{th}$ of its initial value?

A. 20 minutes B. 40 minutes C. 60 minutes D. 80 minutes

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Explanation

$1/16 = (1/2)^4$, so 4 half-lives $= 4\times 20 = 80$ min.

NEET 2024

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${}^{290}_{82}X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$. In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are:

A. 286, 81 B. 280, 81 C. 286, 80 D. 288, 82

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Explanation

$Q = (286, 81)$ after $\alpha$, $\beta^+$, $\beta^-$, $\beta^-$.

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