Question

After a certain lapse of time, the fraction of radioactive polonium left is found to be $12.5$% of initial quantity. The duration of this time lapse is (if the half life of polonium is $138$ days)

• A. $414days$
• B. $407days$
• C. $421days$
• D. $410days$

Answer 1

Answer: (A)

$414days$

$12.5$ of initial quantity i.e. ${\frac{1}{8}}^{th}=(\frac{1}{2}{)}^3$

i.e. after $3$ half life times,

If $t_{1/2}=138days$

Then duration of this time lapse $=3t_{1/2}$

$=3\times 138$

$=414days$