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A radioactive element emits 200 particles per second. After three hours 25 particles per second are emitted. The half life period of element will be

50 minutes

60 minutes

70 minutes

80 minutes

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Explanation

(b) $R = \frac{dN}{dt} \propto N \Rightarrow \frac{R_{2}}{R_{1}} = \frac{N_{2}}{N_{1}}$

But $\frac{N_{2}}{N_{1}} = {(\frac{1}{2})}^{t_{1 / 2}} \Rightarrow \frac{25}{200} = \frac{1}{8} = {(\frac{1}{2})}^{3} \Rightarrow \frac{t}{t_{1 / 2}} = 3$

$∴ t_{1 / 2} = \frac{t}{3} = \frac{3}{3} = 1 hour = 60 minutes$

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The half life of the isotope ${}_{11}^{24}{Na}$ is 15 hrs. How much time does it take for $\frac{7}{8} th$ of a sample of this isotope to decay

75 hrs

65 hrs

55 hrs

45 hrs

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Explanation

(d) Undecayed isotope $= 1 - \frac{7}{8} = \frac{1}{8}$

$∴ \frac{N}{N_{0}} = {(\frac{1}{2})}^{t / T} \Rightarrow (\frac{1}{8}) = {(\frac{1}{2})}^{t / 15} \Rightarrow \frac{t}{15} = 3$

or t = 45 hours

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If the half life of a radioactive sample is 10 hours,
its mean life is

14.4 hours

7.2 hours

20 hours

6.93 hours

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Explanation

(a) $Mean life = \frac{Half life}{0.6931} = \frac{10}{0.6931} = 14.4 hours$

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If 20 gm of a radioactive substance due to radioactive decay reduces to 10 gm in 4 minutes, then in what time 80 gm of the same substance will reduce to 10 gm

In 8 minutes

In 12 minutes

In 16 minutes

In 20 minutes

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Explanation

(b) 20 gm substance reduces to 10 gm (i.e. becomes half in 4 min. So $T_{1 / 2}$ = 4 min . Again

$M = M_{0} {(\frac{1}{2})}^{t / T_{1 / 2}}$

$\Rightarrow 10 = 80 {(\frac{1}{2})}^{t / 4} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{t / 4} \Rightarrow$ t = 12 min.

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16 gm sample of a radioactive element is taken from Bombay to Delhi in 2 hour and it was found that 1 gm of the element remained (undisintegrated). Half life of the element is

2 hour

1 hour

$\frac{1}{2}$ hour

$\frac{1}{4} hour$

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Explanation

(c) $N = N_{0} {(\frac{1}{2})}^{t / T_{1 / 2}} \Rightarrow 1 = 16 {(\frac{1}{2})}^{\frac{2}{T_{1 / 2}}} \Rightarrow T_{1 / 2} = \frac{1}{2} hour$

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Unit of radioactivity is Rutherford. Its value is

$3.7 \times 10^{10}$ disintegrations/sec

$3.7 \times 10^{6}$ disintegrations/sec

$1.0 \times 10^{10}$ disintegrations/sec

$1.0 \times 10^{6}$ disintegrations/sec

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Explanation

(d)

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A radioactive substance has a half life of 60 minutes. After 3 hours, the fraction of atom that have decayed would be

12.5%

87.5%

8.5%

25.1%

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Explanation

(b) $N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$ Hence fraction of atoms decayed
= $1 - \frac{N}{N_{0}} = 1 - {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} = 1 - {(\frac{1}{2})}^{\frac{3 \times 60}{60}} = \frac{7}{8}$
In percentage it is $\frac{7}{8} \times 100 = 87.5 %$

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An element used for radioactive carbon dating for more than 5600 years is

C-14

U-234

U-238

Po-94

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Explanation

(a) C-14 is carbon dating substance.

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N atoms of a radioactive element emit n alpha particles per second. The half life of the element is

$\frac{n}{N}$ sec

$\frac{N}{n}$ sec

$\frac{0.693 N}{n}$ sec

$\frac{0.693 n}{N}$ sec

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Explanation

$∴ λ = \frac{- n}{N}$ Half life = $\frac{0.693}{λ} = \frac{0.693 N}{n} \sec$

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The radioactive nucleus ${}_{7}^{13}N$ decays to ${}_{6}^{13}C$ through the emission of

Neutron

Proton

Electron

Positron

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Explanation

(d) ${}_{7}^{13}N \to {}_{6}^{13}C + {}_{+ 1}^{0}e$

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