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A radioactive decay chain starts from ${}_{93}^{237}{Np}$ and produces ${}_{90}^{229}{Th}$ by successive emissions. The emitted particles can be

Two $α$ -particles and one $β$ -particle

Three $β^{+}$ particles

One $α$ particle and two $β^{+}$ particles

One $α$ particle and two $β^{-}$ particles

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Explanation

(a) By using $n_{α} = \frac{A - A'}{4}$ and $n_{β} = 2 n_{α} - Z + Z'$

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The half-life of a sample of a radioactive substance is 1 hour. If $8 \times 10^{10}$ atoms are present at t = 0, then the number of atoms decayed in the duration t = 2 hour to t = 4 hour will be

$2 \times 10^{10}$

$1.5 \times 10^{10}$

Zero

Infinity

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Explanation

(b) $N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
No of atoms at t = 2hr, $N_{1} = 8 \times 10^{10} {(\frac{1}{2})}^{\frac{2}{1}} = 2 \times 10^{10}$
No. of atoms at t = 4hr, $N_{2} = 8 \times 10^{10} {(\frac{1}{2})}^{\frac{4}{1}} = \frac{1}{2} \times 10^{10}$
$∴$ No. of atoms decayed in given duration

$= (2 - \frac{1}{2}) \times 10^{10} = 1.5 \times 10^{10}$

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Carbon dating is best suited for determining the age of fossils if their age in years is of the order of

$10^{3}$

$10^{4}$

$10^{5}$

$10^{6}$

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Explanation

(b)

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A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is

120 min

80 min

30 min

20 min

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Explanation

(d) $A = A_{0} {(\frac{1}{2})}^{n} \Rightarrow 30 = 240 {(\frac{1}{2})}^{n} \Rightarrow {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n} \Rightarrow n = 3$

$∴ \frac{t}{T_{1 / 2}} = 3 \Rightarrow T_{1 / 2} = \frac{t}{3} = \frac{1}{3} hr = 20 \min$

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The half-life of a radioactive substance is 40 years. How long will it take to reduce to one fourth of its original amount and what is the value of decay constant

40 year, 0.9173/year

90 year, 9.017/year

80 year, 0.0173 year

None of these

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Explanation

(c) To reduce one fourth it takes time $t = 2 (T_{1 / 2})$ = 2 $\times$ 40
= 80 years.
Decay constant $λ = \frac{0.693}{T_{1 / 2}} = \frac{0.693}{40} = 0.0173 years$

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A radioactive material has a half life of 10 days. What fraction of the material would remain after 30 days

0.5

0.25

0.125

0.33

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Explanation

(c) $N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow \frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{30}{10}} = \frac{1}{8} = 0.125$

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Starting with a sample of pure ${}_{66}{Cu}$ , $\frac{7}{8}$ of it decays into Zn in 15 min. The corresponding half-life is

5 min

$7 \frac{1}{2}$ min

10 min

15 min

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Explanation

(a) $\frac{7}{8}$ part decays i.e. remaining part is $\frac{1}{8}$

$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{\frac{15}{T_{1 / 2}}} \Rightarrow T_{1 / 2} = 5 \min$

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A radioactive material has a half-life of 8 years. The activity of the material will decrease to about 1/8 of its original value in

256 year

128 year

64 year

24 year

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Explanation

(d) $\frac{A}{A_{0}} = {(\frac{1}{2})}^{t / T_{1 / 2}} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{t / 8} \Rightarrow t = 24 years$

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A nucleus of mass 218 amu in free state decays to emit an $α$ -particle. Kinetic energy of the $α$ -particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is

1.0

0.5

0.25

0.125

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Explanation

(d) Using conservation of momentum $P_{daughter} = P_{α}$
$\Rightarrow \frac{E_{d}}{E_{α}} = \frac{m_{α}}{m_{d}} \Rightarrow E_{d} = \frac{E_{α} \times m_{α}}{m_{d}} = \frac{6.7 \times 4}{214} = 0.125 MeV$

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$C^{14}$ has half life 5700 years. At the end of 11400 years, the actual amount left is

0.5 of original amount

0.25 of original amount

0.125 of original amount

0.0625 of original amount

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Explanation

(b) $N = N_{0} \times {(\frac{1}{2})}^{11400 / 5700} = N_{0} {(\frac{1}{2})}^{2} = 0.25 N_{0}$

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