Chapter 26, Problem 40P

To determine

The electron’s speed.

Answer to Problem 40P

The speed of an electron is $0.833c$ .

Explanation of Solution

Given:

Rest Mass, $m_0=9.11\times {10}^{-31}kg$

Potential energy decreases by $6.60\times {10}^{-14}J$

Formula used:

It is known that the increase in kinetic energy comes from the decrease in potential energy:

  $KE=\left(m-m_0\right)c^2=m_{0}c\left(\left\{\frac{1}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}\right\}-1\right)$

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:

Substitute the given values

  $6.60\times {10}^{-14}J=\left(9.11\times {10}^{-31}kg\right){\left(3.00\times {10}^{8}m/s\right)}^2\left(\left\{\frac{1}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}\right\}-1\right)$

  $v=0.833c$

Conclusion: The required value is $0.833c$ .