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Physics MCQs for NEET — Practice Questions with Answers

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An atomic nucleus ${}_{90}^{232}{Th}$ emits several $α$ and $β$ radiations and finally reduces to ${}_{82}^{208}{Pb}$ . It must have emitted

4 $α$ and 2 $β$

6 $α$ and 4 $β$

8 $α$ and 24 $β$

4 $α$ and 16 $β$

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Explanation

(b) $n_{α} = \frac{A - A'}{4} = \frac{232 - 208}{4} = 6$

$n_{β} = 2 n_{α} - Z + Z' = 2 \times 6 - 90 + 82 = 4$

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A radioactive substance has a half-life of 1 year. The fraction of this material, that would remain after 5 years will be

$\frac{1}{32}$

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{4}{5}$

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Explanation

(a) Number of half lives $n = \frac{5}{1} = 5$
Now $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{N}{N_{0}} = {(\frac{1}{2})}^{5} = \frac{1}{32}$

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If half-life of a substance is 3.8 days and its quantity is 10.38 gm. Then substance quantity remaining left after 19 days will be

0.151 gm

0.32 gm

1.51 gm

0.16 gm

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Explanation

(b) Number of half lives $n = \frac{19}{3.8} = 5$ ; Now $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n}$

$\Rightarrow \frac{N}{10.38} = {(\frac{1}{2})}^{5} \Rightarrow N = 10.38 \times {(\frac{1}{2})}^{5} = 0.32 gm$

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Which is the correct expression for half-life

$t_{1 / 2} = \log 2$

$t_{1 / 2} = \frac{λ}{\log 2}$

$t_{1 / 2} = \frac{λ}{\log 2} (2.303)$

$t_{1 / 2} = \frac{2.303 \log 2}{λ}$

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Explanation

(d) $T_{1 / 2} = \frac{\log_{e} 2}{λ} = \frac{2.303 \log_{10} 2}{λ}$

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A nucleus ${}_{Z}^{A}A$ emits an $α$ -particle. The resultant nucleus emits a $β^{+}$ particle. The respective atomic and mass no. of the final nucleus will be

Z – 3, A – 4

Z – 1, A – 4

Z – 2, A – 4

Z, A – 2

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Explanation

(a) ${}_{Z}^{A}X \to {}_{Z - 2}^{A - 4}Y + {}_{2}^{4}{He} \to {}_{Z - 3}^{A - 4}Z + {}_{1}^{0}B$

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The activity of a radioactive sample is 1.6 curie and its half-life is 2.5 days. Its activity after 10 days will be

0.8 curie

0.4 curie

0.1 curie

0.16 curie

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Explanation

$\Rightarrow \frac{A}{A_{0}} = \frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow A = 1.6 \times {(\frac{1}{2})}^{4} = 0.1 curie$

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In a mean life of a radioactive sample

About 1/3 of substance disintegrates

About 2/3 of the substance disintegrates

About 90% of the substance disintegrates

Almost all the substance disintegrates

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Explanation

(b) By using $N = N_{0} e^{- λt}$ and $t = τ = \frac{1}{λ}$
Substance remains = $N = \frac{N_{0}}{e} = 0.37 N_{0} ≃ \frac{N_{0}}{3}$
$∴$ Substance disintegrated = $N_{0} - \frac{N_{0}}{3} = \frac{2 N_{0}}{3}$

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Three fourth of the active decays in a radioactive sample in 3/4 sec. The half life of the sample is

$\frac{1}{2}$ sec

1 sec

$\frac{3}{8}$ sec

$\frac{3}{4}$ sec

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Explanation

$t = 2 (T_{1 / 2}) \Rightarrow \frac{3}{4} = 2 (T_{1 / 2}) \Rightarrow T_{1 / 2} = \frac{3}{8} \sec$

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During mean life of a radioactive element, the fraction that disintegrates is

$\frac{1}{e}$

$\frac{e - 1}{e}$

$\frac{e}{e - 1}$

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Explanation

(c) By using $N = N_{0} e^{- λt}$ and average life time $t = \frac{1}{λ}$
So $N = N_{0} e^{- λ \times 1 / λ} = N_{0} e^{- 1} \Rightarrow \frac{N}{N_{0}} = e^{- 1} = \frac{1}{e}$
Now disintegrated fraction = $1 - \frac{N}{N_{0}} = 1 - \frac{1}{e} = \frac{e - 1}{e}$

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1 Curie is equal to

$3 \times 10^{10}$ disintegrations/sec

$3.7 \times 10^{7}$ disintegrations/sec

$5 \times 10^{7}$ disintegrations/sec

$3.7 \times 10^{10}$ disintegrations/sec

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Explanation

(d)

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