A) 4 eV
B) 2 eV
C) 8 eV
D) 6 eV
Correct Answer: C
Solution :
\[r=\frac{\sqrt{2mK}}{qB}\Rightarrow q\propto \sqrt{mK}\Rightarrow K\propto \frac{{{q}^{2}}}{m}\] \[\Rightarrow \frac{{{K}_{\alpha }}}{{{K}_{p}}}={{\left( \frac{{{q}_{\alpha }}}{{{q}_{p}}} \right)}^{2}}\times \frac{{{m}_{p}}}{{{m}_{\alpha }}}\Rightarrow \frac{{{K}_{\alpha }}}{8}={{\left( \frac{2{{q}_{p}}}{{{q}_{p}}} \right)}^{2}}\times \frac{{{m}_{p}}}{4{{m}_{p}}}=1\] \[\Rightarrow {{K}_{\alpha }}=8\ eV\]
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