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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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If half-life of a radioactive atom is 2.3 days, then its decay constant would be

0.1

0.2

0.3

2.3

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Explanation

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A radioactive element ${}_{90}^{238}X$ decay into ${}_{83}^{222}Y$ . The number of $β$ -particles emitted are

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Explanation

(d) Number of $α$ -particles emitted $= \frac{238 - 222}{4} = 4$
This decreases atomic number to 90-4 $\times$ 2 = 82
Since atomic number of ${}_{83}^{222}Y$ is 83, this is possible if one $β$ -particle is emitted.

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A radio isotope has a half life of 75 years. The fraction of the atoms of this material that would decay in 150 years will be

66.6%

85.5%

62.5%

75%

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Explanation

(d) Number of half lives in 150 years $n = \frac{150}{75} = 2$
Fraction of the atom of decayed = $1 - {(\frac{1}{2})}^{n}$

$= 1 - {(\frac{1}{2})}^{2} = \frac{3}{4} = 0.75 \Rightarrow$ Percentage decay = 75%

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An artificial radioactive decay series begins with unstable ${}_{94}^{241}{Pu}$ . The stable nuclide obtained after eight α decays and five $β$ -decays is

${}_{83}^{209}{Bi}$

${}_{82}^{209}{Pb}$

${}_{82}^{205}{Ti}$

${}_{82}^{201}{Hg}$

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Explanation

(a) Mass number decreases by 8 $\times$ 4 = 32
Atomic number decreases by 8 $\times$ 2-5 = 1

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