Chapter 30, Problem 72GP

To determine

To Find: The percentage by which the proton’s mass have to be increased to make the reaction possible.

Answer to Problem 72GP

The percentage by which the proton’s mass have to be increased to make the reaction possible is 0.083%

Explanation of Solution

Given info:

The reaction: ${\mathrm{e}}^{-}+\mathrm{p}\to \mathrm{n}+\mathrm{\nu }$

Formula used:

Q-value of the reaction is given by: $\mathrm{Q}=\mathrm{\Delta }\mathrm{m}{\mathrm{c}}^2$

Calculation:

Due to availability of large number of low-energy electrons, this situation will be catastrophic. According to the reaction, many numbers of protons will turn into neutrons. The weight of neutrons is almost equal to the weight of protons and electrons together in a nucleus. So, this will eliminate the whole chemistry.

The Q-value can be calculated as follows:

  $\begin{array}{l}\mathrm{Q}=\left({\mathrm{m}}_{\mathrm{p}}-{\mathrm{m}}_{\mathrm{n}}\right){\mathrm{c}}^2\\ \mathrm{Q}=\left(1.007825\text{u}-1.008665\text{u}\right){\mathrm{c}}^2\\ \mathrm{Q}=\left(1.007825-1.008665\right){\mathrm{c}}^2\left(931.5\right)\frac{\text{MeV}}{{\mathrm{c}}^2}\\ \mathrm{Q}=-0.782\text{ MeV}\end{array}$

Also, the mass of proton in units of MeV can be given as:

  ${\mathrm{m}}_{\mathrm{p}}=938.272\text{ MeV}$

The percent increase in mass of proton can be given as:

  $\begin{array}{l}=\frac{\mathrm{Q}}{{\mathrm{m}}_{\mathrm{p}}}\times 100\mathrm{\%}\\ =\frac{0.782\text{ MeV}}{938.272\text{ MeV}}\times 100\mathrm{\%}\\ =0.083\mathrm{\%}\end{array}$

Conclusion:

The percentage by which the proton’s mass have to be increased to make the reaction possible is 0.083%