Question

The half-life of radio isotopic bromine - 82 is 36 hours. The fraction which remains after one day is ___________ ×10–2.
(Given antilog 0.2006 = 1.587)

Answer 1

The half-life of bromine-82, $t_{1/2} = 36 \, \text{hours}$.
$t_{1/2} = \frac{0.693}{K}$
$K = \frac{0.693}{36} = 0.01925 \, \text{hr}^{-1}$

For a 1$^\text{st}$-order reaction, the kinetic equation is:
$t = \frac{2.303}{K} \log \frac{a}{a-x}$

For $t = 1 \, \text{day} \, (t = 24 \, \text{hr})$:
$\log \frac{a}{a-x} = \frac{t \times K}{2.303}$
$\log \frac{a}{a-x} = \frac{24 \, \text{hr} \times 0.01925 \, \text{hr}^{-1}}{2.303}$
$\log \frac{a}{a-x} = 0.2006$

Now,
$\frac{a}{a-x} = \text{antilog} \, (0.2006)$

$\frac{a}{a-x} = 1.587$

If $a = 1$:

$\frac{1}{1-x} = 1.587 \implies 1-x = 0.6301$

Thus, the fraction remaining after one day is:

$1-x = 0.6301 \approx 63\%$

The Correct Answer is: 63%