The half-life of radioisotope bromine -82 is 36 hours. The fraction which remains after one day is

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Solution

Q.
The half-life of radioisotope bromine -82 is 36 hours. The fraction which remains after one day is ______ $\times 10^{- 2}$ .

(Given antilog 0.2006 = 1.587) [2024]

Ans.
(63)

For first order reaction,

$k = \frac{2.303}{t} \log_{10} \frac{a_{o}}{a_{t}}$

At half life time i.e 36 hrs:

   $k = \frac{2.303}{36} \log_{10} \frac{a_{o}}{a_{o} / 2}$

   $k = \frac{2.303}{36} \log_{10} 2$ (i)

At after one day i.e 24 hrs:

$k = \frac{2.303}{24} \log_{10} \frac{a_{o}}{a_{(t = 24)}}$ (ii)

Equating (i) and (ii)

$\frac{2.303}{24} \log_{10} \frac{a_{o}}{a_{(t = 24)}} = \frac{2.303}{36} \log_{10} 2$

   $\log_{10} \frac{a_{o}}{a_{(t = 24)}} = \frac{24}{36} \log_{10} 2$

   $\log_{10} \frac{a_{o}}{a_{(t = 24)}} = \frac{24}{36} \times 0.301 = 0.20066$

   $\frac{a_{o}}{a_{(t = 24)}} = antilog (0.2006) = 1.587$

   $\frac{a_{(t = 24)}}{a_{o}} = \frac{1}{1.587} = 0.63 = 63 \times 10^{- 2}$

Thus the fraction which remains after one day is $63 \times 10^{- 2}$ .

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