A proton, a deuteron and an alpha particle, initially at rest, are accelerated through the same potential difference. They then enter a uniform magnetic field perpendicular to their velocities and move in circular paths of radii \( r_p \), \( r_d \) and \( r_\alpha \) respectively. Which one of the following relations is correct?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( r_p\lt r_d=r_\alpha \)
A charge \( q \) of mass \( m \) accelerated from rest through a potential difference \( V \) gains kinetic energy \( qV \). In a uniform magnetic field perpendicular to its velocity, the magnetic force \( qvB \) provides the centripetal force. The radius of the circle is then \( r=mv/qB \).
The momentum gained in the acceleration is found from:
\( \displaystyle \frac{1}{2}mv^{2}=qV \)
\( \Rightarrow\; mv=\sqrt{2mqV} \)
Putting this in the radius:
\( \begin{aligned} r &= \frac{mv}{qB} \\ &= \frac{\sqrt{2mqV}}{qB} \\ &= \frac{1}{B}\sqrt{\frac{2mV}{q}} \end{aligned} \)
With \( V \) and \( B \) the same for all three, \( r\propto\sqrt{m/q} \). Take the proton to have mass \( m \) and charge \( e \). The deuteron then has mass \( 2m \) and charge \( e \). The alpha particle has mass \( 4m \) and charge \( 2e \).
\( r_p\propto\sqrt{m/e} \)
\( r_d\propto\sqrt{2m/e} \)
\( \begin{aligned} r_\alpha &\propto \sqrt{\frac{4m}{2e}} \\ &= \sqrt{2m/e} \end{aligned} \)
From the above, \( r_p:r_d:r_\alpha=1:\sqrt{2}:\sqrt{2} \). Hence, the deuteron and the alpha particle move on circles of the same radius, larger than that of the proton.