Chapter 26, Problem 58GP

To determine

To identify: The particle on the basis of its rest mass.

Answer to Problem 58GP

The particle is electron.

Explanation of Solution

Given:

Speed is $\text{2}{\text{.24×10}}^{\text{7}}\text{m/s}$ .

After the conversion, $\text{v}\text{=}\frac{\left(\text{2}\text{.24}\text{×}{\text{10}}^{\text{7}}\text{m/s}\right)}{\left(\text{3}\text{.00}{\text{×10}}^{\text{8}}\text{m/s}\right)}\text{=}\text{0}\text{.747}\text{c}$

Momentum is $3.07\times {10}^{-22}kg\cdot m/s$

Formula used:

The momentum to calculate the rest mass

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

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

  $p$ is momentum

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

Calculation:

Substitute all the given values:

  $3.07\times {10}^{-22}kg\cdot m/s=\frac{m_0\left(0.747\right)\left(3.00\times {10}^{8}m/s\right)}{{\left[1-{\left(0.747\right)}^2\right]}^{\frac{1}{2}}}$

  $m_0=9.11\times {10}^{-31}kg$

Thus, the particle has a negative charge so it is an electron.

Conclusion: The required particle is electron.