Nuclear Chemistry Ques 12
12. 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 will be
[2001]
(a) $0.08 $ $M$
(b) $0.04 $ $M$
(c) $0.32 $ $M$
(d) $0.16 $ $M$
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Answer:
Correct Answer: 12.(d)
Solution:
- (d) Remaining activity $=0.01 $ $M$
after $24$ $ hrs$
Remaining activity $=$ Initial activity $\times(\frac{1}{2})^{n}$
Used half life time $(n)=\frac{\text{ Total time }}{T _{1 / 2}}=\frac{24}{6}=4$
So, $\quad 0.01=$ Initial activity $\times(\frac{1}{2})^{4}$
Initial activity $=0.01 \times 16=0.16 $ $M$
BETA
