Chapter 32, Problem 47GP

(a)

To determine

To Calculate: The energy released when an electron and a positron annihilate each other.

Answer to Problem 47GP

  $\mathrm{E}=1.022\text{MeV}$

Explanation of Solution

Formula used:

Energy released can be calculated as:

  $\mathrm{E}=\left({\mathrm{M}}_{{\mathrm{e}}^{-}}+{\mathrm{m}}_{{\mathrm{e}}^{+}}\right){\mathrm{c}}^2$

Where,

  ${\mathrm{M}}_{{\mathrm{e}}^{-}}$ = mass of electron

  ${\mathrm{m}}_{{\mathrm{e}}^{+}}$ = mass of positron

  $\mathrm{c}$ = speed of light

Calculation:

  ${\text{e}}_{-1}^0+{\text{e}}_{+1}^0\to \text{energy}$

  $\mathrm{E}=\left({\mathrm{M}}_{{\mathrm{e}}^{-}}+{\mathrm{m}}_{{\mathrm{e}}^{+}}\right){\mathrm{c}}^2$

Plugging the values

  $\begin{array}{l}\mathrm{E}=\left(0.511+0.511\right)\text{MeV}\\ \mathrm{E}\text{ =1}\text{.022}\text{MeV}\end{array}$

Conclusion:

  $\mathrm{E}\text{ =1}\text{.022}\text{MeV}$ is the energy released when an electron and a positron annihilate each other.

(b)

To determine

To Calculate: The energy released when a proton and an antiproton annihilate each other.

Answer to Problem 47GP

  ${\mathrm{E}}_{\mathrm{P}}=\text{1876}\text{.6}\text{MeV}$

Explanation of Solution

Formula used:

Energy released can be calculated as:

  $\mathrm{E}=\left({\mathrm{M}}_{\mathrm{p}}+{\mathrm{m}}_{{\mathrm{p}}^{-}}\right){\mathrm{c}}^2$

Where,

  ${\mathrm{M}}_{\mathrm{p}}$ = mass of proton

  ${\mathrm{m}}_{{\mathrm{p}}^{-}}$ = mass of antiproton

  $\mathrm{c}$ = speed of light

Calculation:

  $\mathrm{p}+\overline{\mathrm{p}}\to \text{energy}$

  $\mathrm{E}=\left({\mathrm{M}}_{\mathrm{p}}+{\mathrm{m}}_{{\mathrm{p}}^{-}}\right){\mathrm{c}}^2$

Plugging the values

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{P}}=\left(938.3\times 2\right)\text{MeV}\\ {\mathrm{E}}_{\mathrm{P}}\text{=1876}\text{.6}\text{MeV}\end{array}$

Conclusion:

  ${\mathrm{E}}_{\mathrm{P}}=\text{1876}\text{.6}\text{MeV}$