Answer
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Work Step by Step
Find the speed of each charged particle using energy conservation. The electrical potential energy becomes kinetic energy.
$$qV=\frac{1}{2}mv^2$$
$$v=\sqrt{\frac{2qV}{m}}$$
When the particle enters the magnetic field, the magnetic force causes it to move in a circular path. Use the relationship from Example 20-6, page 567, and combine it with the above expression.
$$r=\frac{mv}{qB} =\frac{1}{B}\sqrt{\frac{2mV}{q}}$$
We are asked to express the radius of the path for the deuteron in terms of that for the proton.
$$\frac{r_d}{r_p}=\frac{\frac{1}{B}\sqrt{\frac{2m_dV}{q_d}}}{\frac{1}{B}\sqrt{\frac{2m_pV}{q_p}}}$$
$$\frac{r_d}{r_p}=\frac{\sqrt{\frac{m_d}{m_p}}}{\sqrt{\frac{q_d}{q_p}}}=\sqrt{2}$$
$$r_d=\sqrt{2}r_p$$
We are asked to express the radius of the path for the alpha particle in terms of that for the proton.
$$\frac{r_\alpha}{r_p}=\frac{\frac{1}{B}\sqrt{\frac{2m_\alpha V}{q_\alpha }}}{\frac{1}{B}\sqrt{\frac{2m_pV}{q_p}}}$$
$$\frac{r_\alpha }{r_p}=\frac{\sqrt{\frac{m_\alpha }{m_p}}}{\sqrt{\frac{q_\alpha }{q_p}}}=\sqrt{2}$$
$$r_\alpha =\sqrt{2}r_p$$