Question

The activity of a radioisotope falls to 12.5% in 90 days. The half-life of the isotope is $5X$. Then $X$ is:

Answer 1

Given N = 12.5, $N_0=100$
t = 90 days

$K=\frac{2.303}{t}log\frac{N_0}{N}=\frac{2.303}{90}log\frac{100}{12.5}=2.311\times {10}^{-2}day{s}^{-1}$

$t_{1/2}=\frac{0.693}{K}=\frac{0.693}{2.311\times {10}^{-2}}=29.99\approx 30$ days

SInce, $5X=30$

$\therefore X$= $6$.