Question

${}^{60}Co$ has $t_{1/2}=53$ years. The time taken for $7/8$ of the original sample to disintegrate will be:

• A. $4.6$ years
• B. $9.2$ years
• C. $10.6$ years
• D. $15.9$ years

Answer 1

The correct option is C $10.6$ years

Radioactive decay is first order reaction so given $t_{1/2}=53$ years
$k=\frac{0.693}{53}$
$⟹k=1.3\times {10}^{-2}$

Since we know $t=\frac{2.303}{k}log\frac{1}{1-1/8}$

$⟹t=\frac{2.303}{1.3\times {10}^{-2}}log\frac{8}{7}$
$⟹t=10.6$ years