Chapter 26, Problem 23P

To determine

The rest mass of a proton in $MeV/c^2$ .

Answer to Problem 23P

  $938MeV/c^2$ .

Explanation of Solution

Given:

  $1MeV=1.60\times {10}^{-13}J$

Formula used:

It is known that $m_0=\frac{E}{c^2}$

Where,

  $E$ is the energy

  $m_0$ is the rest mass

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

Calculation:

It is known that the mass of a proton is $E=m{c}^2=938MeV$

Substitute the given values in the formula: $m_0=\frac{E}{c^2}$

  $m_0=\frac{\left(1.67\times {10}^{-27}kg\right){\left(2.9979\times {10}^{8}m/s\right)}^2}{\left(1.60\times {10}^{-13}J/MeV\right)c^2}=8.20\times {10}^{-14}J=938MeV/c^2$

Conclusion:

The required value is $938MeV/c^2$ .