Question

A proton and an alpha particle, accelerated through same potential difference, enter in a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them.

Answer 1

Answer:

$\frac{r_p}{r_{\alpha}}=\frac{1}{\sqrt{2}}$

Explanation:

Radius of circular path r of a charge q with mass m and moving with velocity v in a magnetic field B is given by

$r=\frac{mv}{qB}$

If the charge q is accelerated through potential difference V then

$\frac{1}{2}mv^2=qV$

$\implies v=\sqrt{\frac{2qV}{m}}$

Therefore

$r=\frac{m}{qB}\times\sqrt{\frac{2qV}{m}}$

$r=\frac{1}{B}\sqrt{\frac{2mV}{q}}$

Therefore, for the proton

$r_p=\frac{1}{B}\sqrt{\frac{2m_pV}{q_p}}$

And, for the alpha particle

$r_{\alpha}=\frac{1}{B}\sqrt{\frac{2m_{\alpha}V}{q_{\alpha}}}$

Thus,

$\frac{r_p}{r_{\alpha}}=\sqrt{\frac{m_p}{q_p}\times\frac{q_{\alpha}}{m_{\alpha}}}$

We know that

$m_{\alpha}=4m_p$ and $q_{\alpha}=2q_p$

Thus

$\frac{r_p}{r_{\alpha}}=\sqrt{\frac{1}{4}\times\frac{2}{1}}$

$\implies \frac{r_p}{r_{\alpha}}=\frac{1}{\sqrt{2}}$

Hope this helps.