If $20\ g$ of a radioactive substance due to radioactive decay reduces to $10\ g$ in 4 minutes, then in what time $80\ g$ of the same substance will reduce to $10\ g$

Question

Dr. Abubacker · Accepted Answer

(b) $20\ g$ substance reduces to $10\ g$ ($i.e.,$ becomes half in 4 min. So $T_{1/2} = 4$ min. Again $M = M_{0} = \left( \frac{1}{2} \right)^{t/T_{1/2}}$
$\Rightarrow 10 = 80\left( \frac{1}{2} \right)^{t/4} \Rightarrow \frac{1}{8} = \left( \frac{1}{2} \right)^{3} = \left( \frac{1}{2} \right)^{t/4} \Rightarrow t = 12\ min$

Q. If $20\ g$ of a radioactive substance due to radioactive decay reduces to $10\ g$ in 4 minutes, then in what time $80\ g$ of the same substance will reduce to $10\ g$

Explanation

Reviewed by Dr. Abubacker

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