The half-life of radioisotopic bromine -82 is | vedclass

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(C) The radioactive decay follows $1^{st}$ order kinetics.Half-life $t_{1/2} = 36 \ hours$.Decay constant $K = \frac{0.693}{t_{1/2}} = \frac{0.693}{36

Vedclass · Accepted Answer

$63$

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(C) The radioactive decay follows $1^{st}$ order kinetics.Half-life $t_{1/2} = 36 \ hours$.Decay constant $K = \frac{0.693}{t_{1/2}} = \frac{0.693}{36} = 0.01925 \ hr^{-1}$.For $1^{st}$ order reaction,$\ln \frac{N_0}{N_t} = Kt$,where $N_t/N_0$ is the fraction remaining.$\log \frac{N_0}{N_t} = \frac{Kt}{2.303}$.Given $t = 1 \ day = 24 \ hours$.$\log \frac{N_0}{N_t} = \frac{0.01925 	imes 24}{2.303} = \frac{0.462}{2.303} \approx 0.2006$.$\frac{N_0}{N_t} = 	ext{antilog } (0.2006) = 1.587$.Fraction remaining $\frac{N_t}{N_0} = \frac{1}{1.587} \approx 0.6301$.Thus,the fraction remaining is $63.01 	imes 10^{-2} \approx 63 	imes 10^{-2}$.

The half-life of radioisotopic bromine $-82$ is $36 \ hours$. The fraction which remains after one day is . . . . . . $\times 10^{-2}$. (Given $\text{antilog } 0.2006 = 1.587$)

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