Question

Four particles (proton, deuteron, alpha particle and positron) are projected perpendicular to a uniform magnetic field with same momentum. The decreasing order of the radius of curvature of their paths is

• A. positron, proton, alpha particle, deuteron
• B. proton, positron, deuteron, alpha particle
• C. alpha particle, deuteron, positron, proton
• D. positron, alpha particle, deuteron, proton

Answer 1

Answer: (B)

proton, positron, deuteron, alpha particle

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

So, $r\propto \frac{1}{q}$ $(\because \frac{mV}{B}=constant)$

So, the particle having less charge has more radius of curvature.

So, $r_{proton}=r_{po}=r_d>r_{\alpha }$