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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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The activity of a sample is $64 \times 10^{- 5}$ Ci. Its half-life is 3 days. The activity will become $5 \times 10^{- 6}$ Ci after

12 days

7 days

18 days

21 days

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Explanation

(d) $A = A_{0} {(\frac{1}{2})}^{t / T_{1 / 2}} \Rightarrow 5 \times 10^{- 6} = 64 \times 10^{- 5} {(\frac{1}{2})}^{t / 3} \Rightarrow \frac{1}{128} = {(\frac{1}{2})}^{t / 3} \Rightarrow t = 21 days$

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The half-life of radon is 3.8 days. Three forth of a radon sample decay in

5.02 days

15.2 days

7.6 days

11.4 days

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Explanation

Now $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{4} = {(\frac{1}{2})}^{n} \Rightarrow n = 2$

$\Rightarrow t = n \times T_{1 / 2} = 2 \times 3.8 = 7.6 days$

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A nucleus of an element ${}_{84}^{202}X$ emits an $α$ -particle first, a $β$ -particle next and then a gamma photon. The final nucleus formed has an atomic number

200

199

198

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Explanation

${}_{82}^{198}Y \overset{β - decay}{\to} {}_{83}^{198}Z + {}_{- 1}^{0}β$

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Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{8}$

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Explanation

(d) $n = \frac{72000}{24000} = 3; Now \frac{N}{N_{0}} = {(\frac{1}{2})}^{n} = \frac{1}{8}$

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A radioactive substance emits

$α$ -rays

$β$ -rays

$γ$ -rays

All of these

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Explanation

(d) A radioactive substance can emit all of these rays.

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The large scale destruction, that would be caused due to the use of nuclear weapons is called

Nuclear holocaust

Thermo-nuclear reaction

Nuetron reproduction factor

None of these

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Explanation

(a) Nuclear holocast is a large scale destraction that would be caused duee to use of nuclear weapons

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