Answer
$\frac{KE_{\alpha}}{KE_{\beta}}=5.49\times10^{-4}$.
Work Step by Step
Assume that the particles are not relativistic. Use equation 20-12, r = p/qB. If the the radii traveled by the two particles are equal, the momentum of the alpha particle is twice that of the electron, because the charge of the alpha particle is twice that of the electron (considering magnitudes only).
$$p_{\alpha}=2p_{\beta}$$
Calculate the ratio of kinetic energies.
$$\frac{KE_{\alpha}}{KE_{\beta}}=\frac { p_{\alpha}^2/2m_{\alpha}}{ p_{\beta}^2/2m_{\beta}}$$
Use the ratio of the momenta.
$$\frac{KE_{\alpha}}{KE_{\beta}}=\frac {4 p_{\beta}^2/2m_{\alpha}}{ p_{\beta}^2/2m_{\beta}}$$
$$\frac{KE_{\alpha}}{KE_{\beta}}=\frac {4 m_{\beta}}{m_{\alpha}}$$
For the alpha particle mass, subtract the mass of two electrons from helium’s atomic mass.
$$\frac{KE_{\alpha}}{KE_{\beta}}=\frac {4 (0.000549u)}{4.002603u-2(0.000549u)}$$
$$\frac{KE_{\alpha}}{KE_{\beta}}=5.49\times10^{-4}$$
