Question

A deuteron and an alpha particle with the same kinetic energy move in circular paths under the effect of the same magnetic field. The ratio of the radii \((r_d:r_\alpha)\) of their trajectories is

• A. \(1:1\)
• B. \(1:\sqrt{2}\)
• C. \(\sqrt{2}:1\)
• D. \(2:1\)

Hint:

For particles having the same kinetic energy in the same magnetic field: \[ r=\frac{\sqrt{2mK}}{qB} \] Therefore, \[ r\propto \frac{\sqrt{m}}{q} \] Use this shortcut directly in objective questions.

Answer 1

The Correct Option is C

Concept: The radius of the circular path of a charged particle moving perpendicular to a magnetic field is \[ r=\frac{mv}{qB} \] Using \[ K=\frac{1}{2}mv^2 \] we get \[ v=\sqrt{\frac{2K}{m}} \] Substituting in the expression for \(r\), \[ r=\frac{\sqrt{2mK}}{qB} \] Thus, \[ r\propto \frac{\sqrt{m}}{q} \] for particles having the same kinetic energy in the same magnetic field.

Step 1: Write the mass and charge of each particle. For a deuteron, \[ m_d=2u, \qquad q_d=e \] For an alpha particle, \[ m_\alpha=4u, \qquad q_\alpha=2e \]

Step 2: Find the ratio of radii. \[ \frac{r_d}{r_\alpha} = \frac{\sqrt{m_d}/q_d} {\sqrt{m_\alpha}/q_\alpha} \] \[ = \frac{\sqrt{2u}/e} {\sqrt{4u}/(2e)} \] \[ = \frac{\sqrt{2u}}{e} \cdot \frac{2e}{2\sqrt{u}} \] \[ = \frac{\sqrt{2u}}{\sqrt{u}} \] \[ = \sqrt{2} \] Therefore, \[ r_d\propto \frac{\sqrt{2u}}{e} \] \[ r_\alpha\propto \frac{\sqrt{4u}}{2e} = \frac{2\sqrt{u}}{2e} = \frac{\sqrt{u}}{e} \] Therefore, \[ \frac{r_d}{r_\alpha} = \frac{\sqrt{2u}/e}{\sqrt{u}/e} = \sqrt{2} \]

Step 3: State the answer. \[ \boxed{ r_d:r_\alpha = \sqrt{2}:1 } \] Hence, the correct option is \[ \boxed{(C)} \]