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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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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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