Question

The activity of a radioactive isotope falls to $12.5\%$ in 90 days. The half life and decay constant of isotope are :

• A. $2.31\times {10}^{-2}da{y}^{-1}$; 30 days
• B. $2.65\times {10}^{-2}da{y}^{-1}$; 25 days
• C. $2.12\times {10}^{-2}da{y}^{-1}$; 10 days
• D. none of these

Answer 1

The correct option is A $2.31\times {10}^{-2}da{y}^{-1}$; 30 days

$N_0=100$, $N=12.5$, $t=90$ days $K=\frac{2.303}{t}log\frac{N_0}{N}$

$=\frac{2.303}{90}log\frac{100}{12.5}=2.31\times {10}^{-2}day{s}^{-1}$

$t_{1/2}=\frac{0.693}{K}=\frac{0.693}{2.31\times {10}^{-2}}=30days$

Alternate solution

$\frac{N_0}{N}=\frac{100}{12.5}=8$

$N=\frac{N_0}{8}=\frac{N_0}{2^3}=\frac{N_0}{2^n}$

Number of half-lives = 3

$T=t_{1/2}\times n$

$90=t_{1/2}\times 3$

$t_{1/2}=30days$