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Q.

A human body required 0.01 m activity of radioactive substance after 24 hours. Half life of radioactive substance is 6 hours. Then injection of maximum activity of radioactive substance that can be injected

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a

0.04

b

0.32

c

0.08

d

0.16

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{W}{W_{0}} = \frac{1}{2^{n}} \\ n = \frac{t}{t_{1 / 2}} = \frac{24}{6} = 4 \end{array}$

$\begin{array}{l} \frac{0.01}{w_{0}} = \frac{1}{2^{4}} \\ w_{0} = 0.01 \times 16 \\ = 0.16 \end{array}$

$w_{0} =$ intial amount
$w =$ amount left after disintegration

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A human body required 0.01 m activity of radioactive substance after 24 hours. Half life of radioactive substance is 6 hours. Then injection of maximum activity of radioactive substance that can be injected