Nuclear Chemistry - Result Question 4
####6. 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) $0.16 M$
If species $ _a^{b} X$ emits firstly a positron, then two $\alpha$ and two $\beta$ and in last one $\alpha$ and finally converted to species $ _c^{d} Y$, so correct relation is
[2001]
(a) $c=a-5, d=b-12$
(b) $c=a-6, d=b-8$
(c) $c=a-4, d=b-12$
(d) $c=a-5, d=b-8$
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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 $=0.01 \times 16=0.16 M$
