Chapter 32, Problem 17P

To determine

The minimum kinetic energy of the proton (E_p).

Answer to Problem 17P

Solution:

The minimum kinetic energy of the proton (E_p)is 67.5 MeV.

Explanation of Solution

Given Info:

The energy of the proton is 938.3 MeV/c²

Formula Used:

The formula to calculate the energy is,

$\begin{array}{l}{\text{E=Δmc}}^{\text{2}}\\ \text{E=}\left({\text{m}}_{\text{p}}{\text{+m}}_{\text{p}}{\text{-m}}_{{\text{π}}^{\text{o}}}\right){\text{c}}^{\text{2}}\end{array}$

${\text{E=2m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{-m}}_{{\text{π}}^{\text{0}}}{\text{c}}^{\text{2}}$

• E is the energy produce in the reaction.

• c is the velocity of light at free space (${\text{3×10}}^{\text{8}}\text{m/s}$).

• ${\text{m}}_{\text{p}}$ is the mass of the proton (938.3MeV/c²)

• ${\text{m}}_{{\text{π}}^{\text{0}}}$ is the mass of the ${\text{π}}^{\text{0}}$ (135.0Mev/c²)

The required reaction is

$\text{P+P}\to {\text{2P+π}}^{\text{0}}$

To evaluate the energy of each proton, equate twice of the relativistic proton energy to the sum of the two protons and ${\text{π}}^{\text{0}}$ energy.

$\begin{array}{l}{\text{2αm}}_{\text{p}}{\text{c}}^{\text{2}}{\text{=2m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{0}}}{\text{c}}^{\text{2}}\\ \text{α=}\frac{{\text{2m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{0}}}{\text{c}}^{\text{2}}}{{\text{2m}}_{\text{p}}{\text{c}}^{\text{2}}}\end{array}$

So, the kinetic energy of each proton is,

$\begin{array}{l}\text{E=}\left(\text{α-1}\right){\text{m}}_{\text{p}}{\text{c}}^{\text{2}}\\ \text{E=}\left(\frac{{\text{2m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{0}}}{\text{c}}^{\text{2}}}{{\text{2m}}_{\text{p}}{\text{c}}^{\text{2}}}\text{-1}\right){\text{m}}_{\text{p}}{\text{c}}^{\text{2}}\end{array}$

Calculation:

Substitute 938.3MeV/c² for ‘ ${\text{m}}_{\text{p}}$’and 135.0 MeV/c² for ‘ ${\text{m}}_{{\text{π}}^{\text{0}}}$’, the total energy producein the reaction is

$\text{E=}\left(\frac{\text{2×938}\text{.3MeV+135MeV}}{\text{2×938}\text{.3MeV}}\text{-1}\right)\text{×938}\text{.3MeV}$

$\text{E=67}\text{.5MeV}$ (I)

Thus, minimum kinetic energy of the proton (E_p)is 67.5 MeV.