A) \[1:1\]
B) \[1:2\]
C) \[1:4\]
D) \[1:8\]
Correct Answer: A
Solution :
: Radius of the circular path of a charged particle in the magnetic field is \[r=\frac{m\upsilon }{qB}\] Kinetic energy of the charged particle, \[K=\frac{1}{2}m{{\upsilon }^{2}}\]or \[\upsilon =\sqrt{\frac{2K}{m}}\] \[\therefore \] \[r=\frac{m}{qB}\sqrt{\frac{2K}{m}}=\frac{\sqrt{2mK}}{qB}\] \[r\propto \frac{\sqrt{m}}{q}\]For the same value of K and B. \[\therefore \] \[\frac{{{r}_{\alpha }}}{{{r}_{p}}}=\sqrt{\frac{{{m}_{\alpha }}}{{{m}_{p}}}}\times \frac{{{q}_{p}}}{{{q}_{\alpha }}}=\sqrt{\frac{4m}{m}}\times \frac{e}{2e}=1\]
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