Question

${}_{99}{K}^{40}$ consists of $0.012\%$ of potassium in nature. The human body contains $0.35\%$ potassium by weight. The total radioactivity resulting from ${}_{19}{K}^{40}$ decay in a 75 kg human is :
Half life for ${}_{19}{K}^{40}$ is $1.3\times {10}^9$ years.

• A. $4.81\times {10}^{5}dpm$
• B. $3.59\times {10}^{5}dpm$
• C. $6.12\times {10}^{5}dpm$
• D. none of these

Answer 1

The correct option is A $4.81\times {10}^{5}dpm$

The amount of ${}_{19}{K}^{40}$ $=\frac{0.012}{100}\times \frac{0.35}{100}\times 75\times {10}^3$

$=3.15\times {10}^{-2}g$

$=\frac{3.15\times {10}^{-2}\times 6.023\times {10}^{23}}{40}atoms$

$Rate=\lambda \times Numberofatoms$

$=\frac{0.693}{1.3\times {10}^9\times 365\times 24\times 60}\times \frac{3.15\times {10}^{-2}\times 6.023\times {10}^{23}}{40}$

$=4.81\times {10}^{5}dpm$