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Physics MCQs for NEET — Practice Questions with Answers

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${}_{22}{Ne}$ nucleus after absorbing energy decays into two $α$ -particles and an unknown nucleus. The unknown nucleus is

Nitrogen

Carbon

Boron

Oxygen

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Explanation

(b) ${}_{10}^{22}N \to {}_{2}^{4}{He} + {}_{2}^{4}{He} + {}_{6}^{14}X$ ; hence X is carbon

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The half life of ${}_{131}I$ is 8 days. Given a sample of ${}_{131}I$ at time t = 0, we can assert that

No nucleus will decay before t = 4 days

No nucleus will decay before t = 8 days

All nuclei will decay before t = 16 days

A given nucleus may decay at any time after t = 0

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Explanation

(d) Number of nuclei decreases exponentially
$N = N_{0} e^{- λt}$ and Rate of decay $(- \frac{dN}{dt}) = λN$
Therefore, decay process losts up to t = $\infty$ .
Therefore, a given nucleus may decay at any time after t = 0.

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Carbon-14 decays with half-life of about 5,800 years. In a sample of bone, the ratio of carbon-14 to carbon-12 is found to be $\frac{1}{4}$ of what it is in free air. This bone may belong to a period about x centuries ago, where x is nearest to

2 $\times$ 58

58/2

3 $\times$ 58

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Explanation

(a) To becomes $\frac{1}{4}$ th, it requires time of two half lives
i.e., t = 2( $T_{1 / 2}$ ) = 2 $\times$ 5800 = 2 $\times$ 58 centuries

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In the disintegration series ${}_{92}^{238}U \overset{α}{\to} X \overset{β -}{\to} {}_{Z}^{A}Y$
the values of Z and A respectively will be

92, 236

88, 230

90, 234

91, 234

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Explanation

(d) ${}_{92}^{238}U \overset{α}{\to} X \overset{β -}{\to} {}_{91}^{234}Y$

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The particles emitted by radioactive decay are deflected by magnetic field. The particles will be

Protons and $α$ -particles

Electrons, protons and $α$ -particles

Electrons, protons and neutrons

Electrons and $α$ -particles

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Explanation

(d)

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Half life of a radioactive element is 10 days. The time during which quantity remains 1/10 of initial mass will be

100 days

50 days

33 days

16 days

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Explanation

(c) After three half lives (i.e., 30 days) it remains ${(\frac{1}{2})}^{3} = \frac{1}{8}$ so it will remain $\frac{1}{10}$ th, approximately in 33 days

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If half life of a radioactive element is 3 hours, after 9 hours its activity becomes

1/9

1/27

1/6

1/8

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Explanation

(d) By using $A = A_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow \frac{A}{A_{0}} = {(\frac{1}{2})}^{9 / 3} = \frac{1}{8}$

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The nucleus ${}_{48}^{115}{Cd}$ after two successive $β^{-}$ decays will give

${}_{46}^{115}{Pa}$

${}_{49}^{114}{In}$

${}_{50}^{113}{Sn}$

${}_{50}^{115}{Sn}$

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Explanation

(d) ${}_{48}^{115}{Cd} \overset{2 ({}_{- 1}^{0}β)}{\to} {}_{50}^{115}{Sn}$

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At any instant the ratio of the amount of radioactive substances is 2 : 1. If their half lives be respectively 12 and 16 hours, then after two days, what will be the ratio of the substances

1 : 1

2 : 1

1 : 2

1 : 4

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Explanation

(a) Number of half lives in two days four substance 1 and 2 respectively are $n_{1} = \frac{2 \times 24}{12} = 4$ and $n_{2} = \frac{2 \times 24}{1.6} = 3$
By using $N = N_{0} {(\frac{1}{2})}^{n} \Rightarrow \frac{N_{1}}{N_{2}} = \frac{{(N_{0})}_{1}}{{(N_{0})}_{2}} \times \frac{{(\frac{1}{2})}^{n_{1}}}{{(\frac{1}{2})}^{n_{2}}}$

$= \frac{2}{1} \times \frac{{(\frac{1}{2})}^{4}}{{(\frac{1}{2})}^{3}} = \frac{1}{1}$

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Which of the following isotopes is used for the treatment of cancer

$K^{40}$

${Co}^{60}$

${Sr}^{90}$

$I^{131}$

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Explanation

(b) Co-60 can be used to treat cancer. The gamma-rays from isotope Co-60 are used in cobalt-therapy

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