Answer
(a) $1.00$
(b) $0.0233$
Work Step by Step
(a) We know that $r=\frac{P}{qB}$
For electrons $r_e=\frac{P_e}{eB}$....eq(1)
and for the proton $r_p=\frac{P_p}{eB}$.....eq(2)
Dividing eq(1) by eq(2), we have
$\frac{r_e}{r_p}=\frac{\frac{P_e}{eB}}{\frac{P_p}{eB}}=1.00$
(b) We know that
$\frac{1}{2}m_ev_e^2=\frac{1}{2}m_pv_p^2$
but $v=\frac{erB}{m}$
$\frac{1}{2}m_e(\frac{er_eB}{m_e})^2=\frac{1}{2}m_p(\frac{er_pB}{m_p})^2$
This simplifies to:
$\frac{r_e^2}{m_e}=\frac{r_p^2}{m_p}$
$\implies (\frac{r_e}{r_p})^2=\frac{m_e}{m_p}$
$\implies (\frac{r_e}{r_p})=\sqrt{\frac{m_e}{m_p}} $
We plug in the known values to obtain:
$(\frac{r_e}{r_p})=\sqrt{\frac{9.11\times 10^{-31}Kg}{1.673\times 10^{-27}Kg}}=0.0233$
