d
(d) \(r = \frac{{\sqrt {2mK} }}{{qB}} \Rightarrow K \propto \frac{{{q^2}}}{m}\)
\( \Rightarrow \frac{{{K_p}}}{{{K_d}}} = {\left( {\frac{{{q_p}}}{{{q_d}}}} \right)^2} \times \frac{{{m_d}}}{{{m_p}}} = {\left( {\frac{1}{1}} \right)^2} \times \frac{2}{1} = \frac{2}{1}\)
\( \Rightarrow {k_p} = 2 \times 50 = 100\;keV.\)