Question

The ratio of $N/N_o$ after an hour has passed for a radioactive material of half life 60 s is :

• A. $6.82{10}^{-19}$
• B. $8.82{10}^{-19}$
• C. $9.82{10}^{-19}$
• D. $1.82{10}^{-19}$

Answer 1

The correct option is B $8.82{10}^{-19}$

$K=\frac{2.303}{t}log\frac{N_0}{N}$
$=\frac{0.613}{60s^{-1}(t=1\times 60\times 60s}$
$\frac{0.693}{60}=\frac{2.303}{1\times 60\times 60}log\frac{N_0}{N}$
$\therefore \frac{N_0}{N}=8.82\times {10}^{-19}$