A) The \[\alpha -\]particle will be bent in a circular path with a small radius that for the proton
B) The radius of the path of the \[\alpha -\]particle will be greater than that of the proton
C) The \[\alpha -\]particle and the proton will be bent in a circular path with the same radius
D) The \[\alpha -\]particle and the proton will go through the field in a straight line
Correct Answer: C
Solution :
\[r=\frac{\sqrt{2mK}}{qB}i.e.\ r\propto \frac{\sqrt{m}}{q}\] Here kinetic energy K and B are same. \[\therefore \frac{{{r}_{p}}}{{{r}_{\alpha }}}=\frac{\sqrt{{{m}_{p}}}}{\sqrt{{{m}_{\alpha }}}}.\frac{{{q}_{\alpha }}}{{{q}_{p}}}=\frac{\sqrt{{{m}_{p}}}}{\sqrt{4{{m}_{p}}}}.\frac{2{{q}_{p}}}{{{q}_{p}}}=1\]
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