Question

A radioactive isotope has a half-life of 14.8 hr. How much time it will take for the activity to drop to (a) 10 percent, (b) 5 percent of its original:

• A. (a)49.184 h (b) 64 h
• B. (a)29.184 h (b) 44 h
• C. (a)69.184 h (b) 34 h
• D. None of these

Answer 1

The correct option is A (a)49.184 h (b) 64 h

$\lambda =\frac{2.303}{t}log\frac{C_0}{C}$
$t=\frac{2.303}{\lambda }long\frac{C_0}{C}$
$=\frac{2.303\times t_{50}}{0.693}log\left(\frac{C_0}{C}\right)$
(a) $t=\frac{2.303\times 14.8}{0.693}log\frac{100}{10}$
$=49.184h.$
$t=\frac{2.303\times 14.8}{0.693}log\frac{100}{5}=64h$