Question

The activity of a radioactive sample is measured as $N_0$ counts per minute at $t=0$ and $N_0/e$ counts per minute at $t=5$ min. The time, (in minute) at which the activity reduces to half its value, is :

• A. ${\log }_e\frac{2}{5}$
• B. $\frac{5}{{\log }_{e}2}$
• C. $5{\log }_{10}2$
• D. $5{\log }_{e}2$

Answer 1

Using first order equation $N=N_o{e}^{-\lambda t}$
Given : $N=\frac{N_o}{e}$ @ $t=5min$
$\therefore$ $\frac{N_o}{e}=N_o{e}^{-\lambda \left(5\right)}$
Or $-{\log }_{e}e=-5\lambda$
$⟹$ $\lambda =\frac{1}{5}$ $(\because {\log }_{e}e=1)$
Time at which the activity reduces to half of its initial value is known as half life of the radioactive substance.
Half life $T_{1/2}=\frac{{\log }_{e}2}{\lambda }=5{\log }_{e}2$