Chapter 30, Problem 62GP

To determine

To Find: Decay time of one proton in body.

Answer to Problem 62GP

The decay time of one proton in the body is $\mathrm{\Delta }\mathrm{t}=6.2\times {10}^4\text{ yr}$ .

Explanation of Solution

Given info:

Half-life of proton, ${\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\ge {10}^{32}\text{yr}$

Known data:

Mass of water in human body, $\mathrm{m}=70\text{ kg}=70\times {10}^3\text{g}$

Molar mass of water, $\mathrm{M}=18\text{ g/mol}$

Avogadro’s number, ${\mathrm{N}}_0=6.02\times {10}^{23}\text{atoms/mol}$

Formula used:

Number of molecules can be given as:

  $\mathrm{N}=\frac{\mathrm{m}}{\mathrm{M}}{\mathrm{N}}_0$

Here,

  $\mathrm{m}$ is the mass of molecule

  $\mathrm{M}$ is the molar mass

  ${\mathrm{N}}_0$ is the Avogadro’s number

Decay time can be given as:

  $\frac{1}{\mathrm{\Delta }\mathrm{t}}=\frac{0.693\mathrm{N}}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Here,

  $\mathrm{\Delta }\mathrm{t}$ is the decay time

  ${\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the half-life

  $\mathrm{N}$ is the number of molecules

Calculation:

Firstly, calculate the number of water molecules as follows:

  $\begin{array}{l}\mathrm{N}=\frac{\mathrm{m}{\mathrm{N}}_0}{\mathrm{M}}\\ \mathrm{N}=\frac{\left(70\times {10}^3\text{g}\right)\left(6.02\times {10}^{23}\text{atoms/mol}\right)}{18\text{ g/mol}}\\ \mathrm{N}=2.34\times {10}^{27}\end{array}$

Now, the number of protons in a water molecule is 10. So, decay time can be given as:

  $\begin{array}{l}\frac{1}{\mathrm{\Delta }\mathrm{t}}=\frac{0.693\mathrm{N}}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \mathrm{\Delta }\mathrm{t}=\frac{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{0.693\mathrm{N}}\\ \mathrm{\Delta }\mathrm{t}=\frac{\left({10}^{32}\text{yr}\right)}{\left(0.693\right)\left(2.34\times {10}^{27}\right)}\\ \mathrm{\Delta }\mathrm{t}=6.2\times {10}^4\text{ yr}\end{array}$

Conclusion:

Thus, the decay time of one proton in the body is $\mathrm{\Delta }\mathrm{t}=6.2\times {10}^4\text{ yr}$ .