Question 418: Atoms & Nuclei

Question

A radioactive element has rate of disintegration 9000 disintegration per minute at a particular instant. After two minutes it becomes 3000 disintegration per minute. The decay constant per minute is

Accepted Answer

We know that, $\mathrm{N}(\mathrm{t})=\mathrm{N}_0 \mathrm{e}^{-\lambda s}$
In 2 mins the activity reduces from 9000 d.p.m. to 3000 d.p.m. ,
$$ \begin{aligned} & 3000=9000 \mathrm{e}^{-2 \lambda} \ & \frac{3000}{9000}=\mathrm{e}^{-2 \lambda} \ & \frac{1}{3}=\mathrm{e}^{-2 \lambda} \end{aligned} $$
Taking $\log _e$ on both sides,
$$ \begin{aligned} & \log _e\left(\frac{1}{3}ight)=-2 \lambda \ & -\log _e(3)=-2 \lambda \ & \lambda=\frac{1}{2} \log _e(3)=0.5 \log _e 3 \end{aligned} $$

Answer

$0.5 \log _{\mathrm{c}} 3$

Answer

$0.2 \log _{\mathrm{e}} 3$

Answer

$0.5 \log _{\mathrm{e}} 2$

Answer

$0.2 \log _{\mathrm{e}} 2$

Atoms & Nuclei

A radioactive element has rate of disintegration 9000 disintegration per minute at a particular instant. After two minutes it becomes 3000 disintegration per minute. The decay constant per minute is

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