Question

At a given instant there are 25 % undecayed radioactive nuclei in a sample. After 10 sec the number of undecayed nuclei remains 12.5 %. Calculate :
( i ) mean $-$ life of the nuclei and
( ii ) The time in which the number of undecayed nuclears will further reduce to 6.25 % of the reduced number.

• A. ( i )$t_{means}=14.43s$ ( ii ) 40 seconds
• B. ( i )$t_{means}=1.443s$ ( ii ) 40 seconds
• C. ( i )$t_{means}=14.43s$ ( ii ) 4 seconds
• D. None of these

Answer 1

The correct option is A ( i )$t_{means}=14.43s$ ( ii ) 40 seconds

(i) In 10 seconds, the number of undecayed nuclei decreases from 25% to 12.5%.
Thus, it reduces to one half.
Hence, 10 seconds represent half life period
$t_{1/2}=10s$
The decay constant $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{10}=0.0693/s$
The mean life of the nuclei is $t_{mean}=\frac{1}{\lambda }=\frac{1}{0.0693}=14.43s$
Hence, the mean life of the nuclei is 14.43 s.
(ii) The number of undecayed nuclei further reduces to 6.25 % (or one sixteenth) of the reduced nuclei.
$\frac{100}{16}=6.25$
$\frac{1}{16}=\frac{1}{2^4}$
Thus 4 half life periods or $4\times 10=40s$ will be required.