Chapter 30, Problem 73GP

To determine

To find:The ratio of the kinetic energies for an alpha particle and a beta particle.

Explanation of Solution

Introduction

We assume that the particles are not relativistic, so that $p=\sqrt{2m\text{KE}}$ .

The radius is charge particle rotating in magnetic field is $r=\frac{mv}{qB}.$

If both have same radii

  $\begin{array}{l}\frac{m_{\alpha }{v}_{\alpha }}{2eB}=\frac{m_{\beta }{v}_{\beta }}{eB}\\ m_{\alpha }{v}_{\alpha }=2m_{\beta }{v}_{\beta }\\ p_{\alpha }=2p_{\beta }\end{array}$

Explanation:

The ratio of the kinetic energies for an alpha particle and a beta particle.

  $\frac{{\mathrm{KE}}_{\alpha }}{{\text{KE}}_{\beta }}=\frac{\frac{p_{\alpha }^2}{2m_{\alpha }}}{\frac{p_{\beta }^2}{2m_{\beta }}}=\frac{\frac{4p_{\beta }^2}{2m_{\alpha }}}{\frac{p_{\beta }^2}{2m_{\beta }}}=\frac{4m_{\beta }}{m_{\alpha }}=\frac{4(0.000549\text{u})}{4.002603\text{u}-2(0.000549\text{u})}=5.48\times {10}^{-4}$

Conclusion:

The ratio of the kinetic energies is $5.48\times {10}^{-4}$