Question
An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $${r_e},{r_p},{r_\alpha }$$ respectively in a uniform magnetic field $$B.$$ The relation between $${r_e},{r_p},{r_\alpha }$$ is:
A.
$${r_e} > {r_p} = {r_\alpha }$$
B.
$${r_e} < {r_p} = {r_\alpha }$$
C.
$${r_e} < {r_p} < {r_\alpha }$$
D.
$${r_e} < {r_\alpha } < {r_p}$$
Answer :
$${r_e} < {r_p} = {r_\alpha }$$
Solution :
As we know, radius of circular path in magnetic field
$$r = \frac{{\sqrt {2Km} }}{{qB}}$$
For electron, $${r_e} = \frac{{\sqrt {2K{m_e}} }}{{eB}}\,......\left( {\text{i}} \right)$$
For proton, $${r_p} = \frac{{\sqrt {2K{m_p}} }}{{eB}}\,......\left( {{\text{ii}}} \right)$$
For $$\alpha $$ particle,
$$\eqalign{
& {r_\alpha } = \frac{{\sqrt {2K{m_a}} }}{{{q_\alpha }B}} = \frac{{\sqrt {2K4{m_p}} }}{{2eB}} = \frac{{\sqrt {2K{m_p}} }}{{eB}}\,......\left( {{\text{iii}}} \right) \cr
& \therefore {r_e} < {r_p} = {r_\alpha }\,\,\,\,\,\left( {\because {m_e} < {m_p}} \right) \cr} $$