Question

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

• A. $4.81\times {10}^{5}dpm$
• B. $6.33\times {10}^{5}dpm$
• C. $9.81\times {10}^{5}dpm$
• D. none of these

Answer 1

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

Total mass of ${}_{19}^{40}\text{K}$ $=\frac{0.012}{100}\times \frac{0.35}{100}\times 75\times {10}^3=3.15\times {10}^{-2}g$

Total number of atoms of ${}_{19}^{40}\text{K}$$=\frac{3.15\times {10}^{-2}\times 6.023\times {10}^{23}}{40}=4.74\times {10}^{20}atoms$

The expression for the rate of decay is
$Rate=\lambda \times No.ofatoms$
$=\frac{0.693}{1.3\times {10}^9\times 365\times 24\times 60}\times 4.74\times {10}^{20}=4.81\times {10}^{5}dpm$
Thus, the total radioactivity resulting from ${}_{19}^{40}\text{K}$ decay in a 75 kg human is $4.81\times {10}^{5}dpm$.