A) 25 keV
B) 50 keV
C) 200 keV
D) 100 keV
Correct Answer: D
Solution :
\[r=\frac{\sqrt{2mK}}{qB}\Rightarrow K\propto \frac{{{q}^{2}}}{m}\] \[\Rightarrow \frac{{{K}_{p}}}{{{K}_{d}}}={{\left( \frac{{{q}_{p}}}{{{q}_{d}}} \right)}^{2}}\times \frac{{{m}_{d}}}{{{m}_{p}}}={{\left( \frac{1}{1} \right)}^{2}}\times \frac{2}{1}=\frac{2}{1}\] \[\Rightarrow {{k}_{p}}=2\times 50=100\ keV.\]
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