Question

$\text{K}_{99}^{40}$ consist of 0.012% of potassium in nature. The human body contains 0.35% potassium by weight. The total reactivity resulting from $\text{K}_{19}^{40}$ decay in a 75 kg human is:
Half-life of $\text{K}_{19}^{40}\ is \text{1}\text{.3 }\times \text{ 1}{{\text{0}}^{9}}$ years.
A. $4.81\text{ }\times \text{ 1}{{\text{0}}^{5}}\text{ dpm}$
B. $\text{3}.59\text{ }\times \text{ 1}{{\text{0}}^{5}}\text{ dpm}$
C. $\text{6}.12\text{ }\times \text{ 1}{{\text{0}}^{5}}\text{ dpm}$
D. none of these

Answer 1

Hint: The time required for a quantity to reduce its value up to half as compared to the initial value. The formula for calculating the rate constant from half-life for the first-order reaction will be: $\text{K = }\frac{0.693}{{{\text{t}}_{1/2}}}\,$.
Here, K is the rate constant and ${{\text{t}}_{1/2}}$ is the half-life period of a substance.

Complete answer:
-It is given that the half-life of potassium is $\text{1}\text{.3 }\times \text{ 1}{{\text{0}}^{9}}\text{ years}$, we can calculate the rate constant (k).
-As we know that radioactive decay follows first-order of kinetics, so the formula of the rate constant for the first order will be $\text{K = }\dfrac{0.693}{{{\text{t}}_{1/2}}}\,$. So, $\text{K = }\dfrac{0.693}{\text{1}\text{.3 }\times \text{ 1}{{\text{0}}^{9}}}\ \,\text{= }\,5.3\text{ }\times \,\text{ 1}{{\text{0}}^{-8}}$.

 -Now, we have to calculate the amount of K in the human body whose weight is 75 kg.
-As it is given that in nature and the human body, the amount of K is 0.35% and 0.012% respectively.
-So, the amount of K in 75 kg or $75\text{ }\times \,\text{ 1}{{\text{0}}^{3}}$ gram human body will be: $\text{K = }\dfrac{0.35}{\text{100}}\text{ }\times \text{ }\dfrac{0.012}{100}\text{ }\times \text{ }\,75\text{ }\times \,\text{ 1}{{\text{0}}^{3}}\text{ = 0}\text{.0315g}$.

-Now, to calculate the no. of molecules of \[\text{K}_{19}^{40}\] in 0.0315g, we will use the formula:
$\dfrac{\text{No}\text{. of molecules}}{6.022\text{ }\,\times \text{ 1}{{\text{0}}^{23}}}\text{ = }\dfrac{\text{Mass}}{\text{Molar mass}}$
$\text{No}\text{. of molecules}\text{ = }\frac{0.0315}{40}\text{ }\ \times \text{ }6.022\text{ }\,\times \text{ 1}{{\text{0}}^{23}}\text{ = 0}\text{.47 }\times \text{ 1}{{\text{0}}^{21}}$.

-Now, to calculate the total radioactive we will use the formula of the rate of decay i.e. $\text{R = }\lambda \text{ }\times \text{ N}$.
-Here, R is the rate constant, $\lambda $ is the decay constant and N is the no. of atoms.
-As we have calculated the value of $\lambda $ i.e. $\text{5}\text{.3 }\times \text{ 1}{{\text{0}}^{-8}}\ \text{Years}$.
-So, $\text{R = 5}\text{.3 }\times \text{ 1}{{\text{0}}^{-8}}\ \times \text{ 0}\text{.47 }\times \text{ 1}{{\text{0}}^{21}}\ \text{Years = 2}\text{.49 }\times \text{ 1}{{\text{0}}^{13}}\ \text{Years}$.

Note: First order reaction are those reactions in which the rate of reaction (K) is directly proportional to the concentration of the reactants. In decay, the radioactive atoms decay per unit is directly proportional to the total no. of atoms that’s why it is a first-order reaction.