A deutron of kinetic energy $50\, keV$ is describing a circular orbit of radius $0.5$ $metre$ in a plane perpendicular to magnetic field $\overrightarrow B $. The kinetic energy of the proton that describes a circular orbit of radius $0.5$ $metre$ in the same plane with the same $\overrightarrow B $ is........$keV$
AIPMT 1991, Medium
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(d) $r = \frac{{\sqrt {2mK} }}{{qB}} \Rightarrow K \propto \frac{{{q^2}}}{m}$
$ \Rightarrow \frac{{{K_p}}}{{{K_d}}} = {\left( {\frac{{{q_p}}}{{{q_d}}}} \right)^2} \times \frac{{{m_d}}}{{{m_p}}} = {\left( {\frac{1}{1}} \right)^2} \times \frac{2}{1} = \frac{2}{1}$
$ \Rightarrow {k_p} = 2 \times 50 = 100\;keV.$
$ \Rightarrow \frac{{{K_p}}}{{{K_d}}} = {\left( {\frac{{{q_p}}}{{{q_d}}}} \right)^2} \times \frac{{{m_d}}}{{{m_p}}} = {\left( {\frac{1}{1}} \right)^2} \times \frac{2}{1} = \frac{2}{1}$
$ \Rightarrow {k_p} = 2 \times 50 = 100\;keV.$
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