Answer
$42.8$
Work Step by Step
We know that:
$KE=\frac{1}{2}mv^2$
This simplifies to
$v=\sqrt{\frac{2KE}{m}}$........ eq(1)
We also know that
$r=\frac{mv}{qB}$
Substituting value of "v" from equation (1), we obtain:
$r=\frac{m\sqrt{\frac{2KE}{m}}}{qB}$
$r=\frac{\sqrt{2(KE)m}}{qB}$
Now we can find the ratio of radii of proton and electron
$\frac{r_p}{r_e}=\frac{\frac{\sqrt{2(KE)m_p}}{qB}}{\frac{\sqrt{2(KE)m_e}}{qB}}$
This simplifies to
$\frac{r_p}{r_e}=\sqrt{\frac{m_p}{m_e}}$
We plug in the known values to obtain:
$\frac{r_p}{r_e}=\sqrt{\frac{1.67\times 10^{-27}}{9.11\times 10^{-31}}}=42.8$
