Chapter 26, Problem 36P

To determine

The speed and the momentum of an electron.

Answer to Problem 36P

The speed of an electron is $0.866c$ .

The momentum of an electron is $4.73\times {10}^{-22}kg\cdot m/s$ .

Explanation of Solution

Given:

Mass, $m=9.11\times {10}^{-31}kg$

Kinetic energy is equal to its rest energy

Formula used:

It is known that,

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

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

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

It is known that the momentum of proton or particle:

  $p=\frac{m_{0}v}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$ or $p=\frac{m_{0}v}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Where,

  $m_0$ is rest mass

  $v$ is velocity

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

Calculation:

The total energy of an electron is

  $E=m{c}^2=KE+m_0{c}^2=m_0{c}^2+m_0{c}^2=2m_0{c}^2$

  $m=2m_0$

So, the speed will be $m=\frac{m_0}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

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

  $v=0.866c$

The proton’s momentum is found using $p=\frac{m_{0}v}{{\left[1-\frac{v^2}{c^2}\right]}^{\frac{1}{2}}}$

  $=\frac{\left(9.11\times {10}^{-31}kg\right)\left(0.866\right)\left(3.00\times {10}^{8}m/s\right)}{\left(\frac{1}{2}\right)}$

  $=4.73\times {10}^{-22}kg\cdot m/s$

Conclusion:

The speed is $0.866c$ and momentum is $4.73\times {10}^{-22}kg\cdot m/s$ .