Question

Naturally occurring potassium contains $1.0\times {10}^{-2}$ % of $K^{40}.$ The human body contains $3.0\times {10}^{-1}$ % of K, by weight. The total radioactivity resulting from $K^{40}$ decay from a man of 70 Kg will be: ($t_{1/2}$ for $K^{40}$ is $6.9\times {10}^{14}mi{n}^{-1}$)

• A. $31.6\times {10}^{15}dpm$
• B. $3.16\times {10}^{5}dpm$
• C. $3.16\times {10}^{15}dps$
• D. $31.6\times {10}^{15}dps$

Answer 1

The correct option is B $3.16\times {10}^{5}dpm$

Total number of $K^{40}$ atoms present in a man weighing 70,

$=\frac{1.0\times {10}^{-2}}{100}\times \frac{3.0\times {10}^{-1}}{100\times 40}\times 70\times {10}^3\times 6.02\times {10}^{23}$

$=\frac{1264.2\times {10}^{23}}{4\times {10}^5}=316\times {10}^{18}=3.16\times {10}^{20}atoms.$

The expression for lambda is $\lambda =\frac{0.693}{6.9\times {10}^{14}}=1.004\times {10}^{-15}\simeq {10}^{-15}$

Total radioactivity $A=N\lambda =3.16\times {10}^{20}\times {10}^{-15}=3.16\times {10}^{5}dpm$.