Physics | TS EAMCET 2018 | Discussion Page

An active nucleus decays to $(\frac{1}{3})$ rd in 20 h . tHe fraction of original activity remaining after 80 h is

$\frac{1}{16}$

$\frac{1}{81}$

$\frac{1}{36}$

$\frac{1}{54}$

Option B

Decay equation , $N=N_{0}exp(-\lambda t)$

Given, $N_{1}= N_{0}/3, N_{2}= ?$

$t_{1}=20 h$ , $t_{2}= 20 h$

So, $N_{1}=N_{0}exp(-\lambda t_{1})$

$\Rightarrow$ $\frac{1}{3}=e(-20 \lambda)\Rightarrow e^{20 \lambda}=3$

Now, $N_{2}=N_{0}exp(-\lambda t_{2})$ $\Rightarrow$ $\frac{N_{0}}{N_{2}}= e^{80 \lambda}$

$\Rightarrow$ $\frac{N_{0}}{N_{2}}=(e^{20 \lambda})^{4}=(3)^{4}$

$\Rightarrow$ $N_{2}= \frac{1}{(3)^{4}}N_{0}= \frac{N_{0}}{81}$

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