The selling rate of a radioactive isotope is decided by its activity. The second-hand rate (in rupees) of a one month old $^{32} P$ sample having $t_{1 / 2} = 14.3$ day if it was originally purchased for 800 rupees will be:

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Solution

$k = \frac{0.693}{t_{1 / 2}} = \frac{2.303}{t} l o g \frac{a}{a - x}$

Substitute values in the above expression.

$k = \frac{0.693}{14.3} = \frac{2.303}{30} l o g \frac{800}{a - x}$

$l o g \frac{800}{a - x} = 0.6313$

$\frac{800}{a - x} = 4.2785$
$a - x = 187$

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The selling rate of a radioactive isotope is decided by its activity. The second-hand rate (in rupees) of a one month old 32P sample having t1/2=14.3 day if it was originally purchased for 800 rupees will be:

k=0.693t1/2=2.303tlogaa−xSubstitute values in the above expression.k=0.69314.3=2.30330log800a−xlog800a−x=0.6313800a−x=4.2785a−x=187

The selling rate of a radioactive isotope is decided by its activity. The second-hand rate (in rupees) of a one month old $^{32} P$ sample having $t_{1 / 2} = 14.3$ day if it was originally purchased for 800 rupees will be:

$k = \frac{0.693}{t_{1 / 2}} = \frac{2.303}{t} l o g \frac{a}{a - x}$

Substitute values in the above expression.

$k = \frac{0.693}{14.3} = \frac{2.303}{30} l o g \frac{800}{a - x}$

$l o g \frac{800}{a - x} = 0.6313$

$\frac{800}{a - x} = 4.2785$
$a - x = 187$