A) 40 MeV
B) 80 MeV
C) 20 MeV
D) 60 MeV
Correct Answer: B
Solution :
\[R\,\,=\,\,\frac{\sqrt{2mK}}{qB}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \,\,\,{{R}_{d}}\,=\,{{R}_{p}}\] \[\therefore \,\,\frac{2{{m}_{d}}{{K}_{d}}}{{{q}_{d}}B}\,\,=\,\,\frac{\sqrt{2{{m}_{p}}{{K}_{p}}}}{{{q}_{p}}B}\] \[or\,\,\,\,\frac{\sqrt{2(2{{m}_{p}}){{K}_{d}}}}{eB}\,\,=\,\,\frac{\sqrt{2{{m}_{p}}{{K}_{p}}}}{eB}\] \[\therefore \,\,\text{ }{{K}_{p}}=2{{K}_{d}}\,\text{=}\,2\times 40\text{ }MeV\,\,=\,\,80\text{ }MeV\]
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