Chapter 26, Problem 55GP

To determine

The electromagnetic energy produced.

Answer to Problem 55GP

  $234MeV$

Explanation of Solution

Given:

Rest mass of first nucleon, $m_0=105.7\frac{MeV}{c^2}$

Rest mass of second nucleon, $m_0=105.7\frac{MeV}{c^2}$

Speed of first nucleon, $v_1=0.33c$

Speed of second nucleon, $v_2=0.50c$

Formula used:

It is known that,

  $m=\frac{m_0}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$ or $m=\frac{m_0}{{\left[1-\frac{v^2}{c^2}\right]}^{\frac{1}{2}}}$

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

According to the question, the total energy of the nucleon becomes electromagnetic energy. So,

  $\text{E}\text{=}{\text{m}}_1{c}^2+m_2{c}^2$

  $=\frac{m_0}{{\left[1-\frac{v_{{}^1}^2}{c^2}\right]}^{\frac{1}{2}}}\left(c^2\right)+\frac{m_0}{{\left[1-\frac{v_{{}^2}^2}{c^2}\right]}^{\frac{1}{2}}}\left(c^2\right)$

By substituting all the values:

  $=\frac{\left(\frac{105.7MeV}{c^2}\right)\left(c^2\right)}{{\left[1-{\left(0.33\right)}^2\right]}^{\frac{1}{2}}}+\frac{\left(\frac{105.7MeV}{c^2}\right)\left(c^2\right)}{{\left[1-{\left(0.50\right)}^2\right]}^{\frac{1}{2}}}$

   $=234MeV$

Conclusion: The required value is $234MeV$ .