Chapter 30, Problem 64GP

To determine

To Calculate: The mass of ${}_{19}^{40}\text{K}$ and ${}_{19}^{40}\text{K}$ in 1 L of milk.

Explanation of Solution

Given:

The activity of ${}_{19}^{40}\text{K}$ is given 60 decays/second in 1 L of the milk. From the appendix B, half-life of ${}_{19}^{40}\text{K}$ is $1.277\times {10}^9\text{ years}$ and the percentage abundance of ${}_{19}^{40}\text{K}$ and ${}_{19}^{40}\text{K}$ is 0.0117% and 93.258% respectively.

Formula Used:

Activity is related to decay constant as follows:

  $A=\lambda N$  .......(1)

Here, N is number of atoms.

For radioactive decay, order of the reaction is first order thus,

  $\lambda =\frac{0.6932}{t_{1/2}}$  .......(2)

Here, $t_{1/2}$ is half-life time.

Calculations:

Combining equation (1) and (2),

  $A=\frac{0.6932}{t_{1/2}}\times N$

Putting the values,

  $\frac{60\text{ decay}}{\mathrm{s}}=\frac{0.6932\times N}{1.277\times {10}^9\text{ year}}$

Or,

  $\begin{array}{c}\frac{60\text{ decay}}{\mathrm{s}}=\frac{0.6932\times N}{1.277\times {10}^9\text{ year}}\left(\frac{1\text{ year}}{365\text{ days}}\right)\left(\frac{1\text{ day}}{24\text{ h}}\right)\left(\frac{1\text{ h}}{60\text{ min}}\right)\left(\frac{1\text{ min}}{60\text{ s}}\right)\\ 60=1.721\times {10}^{-17}\times N\\ N=\frac{60}{1.721\times {10}^{-17}}\\ =3.486\times {10}^{18}\end{array}$

Thus, the number of ${}_{19}^{40}\text{K}$ atoms is $3.486\times {10}^{18}$ .

For 1 mol, there are $6.022\times {10}^{23}$ number of atoms thus, number of moles corresponding to $3.486\times {10}^{18}$ number of atoms will be:

  $\begin{array}{c}n=\frac{3.486\times {10}^{18}}{6.022\times {10}^{23}}\text{ mol}\\ =0.5788\times {10}^{-5}\text{ mol}\\ =5.8\times {10}^{-6}\text{ mol}\end{array}$

Molar mass of ${}_{19}^{40}\text{K}$ is 40 g/mol thus, mass can be calculated as follows:

  $\begin{array}{c}m=n\times M\\ =\left(5.8\times {10}^{-6}\right)\left(40\text{ g/mol}\right)\\ =2.32\times {10}^{-4}\text{ g}\end{array}$

Therefore, the mass of ${}_{19}^{40}\text{K}$ in 1 L of the milk is $2.32\times {10}^{-4}\text{ g}$ . The percentage abundance of ${}_{19}^{40}\text{K}$ is 0.0117% so, mass of 0.0117% of ${}_{19}^{40}\text{K}$ is $2.32\times {10}^{-4}\text{ g}$ .

Now, the mass of ${}_{19}^{39}\text{K}$ with percentage abundance 93.258% can be calculated as follows:

  $\begin{array}{c}m=\frac{2.32\times {10}^{-4}\text{ g}}{\frac{0.0117}{100}}\times \frac{93.258}{100}\\ =1.85\text{ g}\end{array}$

Therefore, the mass of ${}_{19}^{39}\text{K}$ in 1 L milk is 1.85 g.

Conclusion:

Thus, the mass of ${}_{19}^{40}\text{K}$ and ${}_{19}^{39}\text{K}$ in 1 L milk is $2.32\times {10}^{-4}\text{ g}$ and 1.85 g respectively.