A) 1 MeV
B) 4 MeV
C) 2 MeV
D) 0.5 MeV
Correct Answer: A
Solution :
\[r=\frac{\sqrt{2mK}}{qB}\] Þ \[K\propto \frac{{{q}^{2}}}{m}\] Þ \[\frac{{{K}_{p}}}{{{K}_{\alpha }}}={{\left( \frac{{{q}_{p}}}{{{q}_{\alpha }}} \right)}^{2}}\times \frac{{{m}_{\alpha }}}{{{m}_{p}}}\] Þ \[\frac{1}{{{K}_{\alpha }}}={{\left( \frac{{{q}_{p}}}{2{{q}_{p}}} \right)}^{2}}\times \frac{4{{m}_{p}}}{{{m}_{p}}}=1\] Þ \[{{K}_{\alpha }}=1\,MeV\].
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