A) \[1meV\]
B) \[4meV\]
C) \[4meV\]
D) \[0.5meV\]
Correct Answer: A
Solution :
As \[r=\frac{\sqrt{2mK}}{qB}\] \[\Rightarrow \] \[K\propto \frac{{{q}^{2}}}{m}\] Suppose the values of K for photon and a-particle are \[{{K}_{P}}\] and \[{{K}_{\alpha }},\] respectively. Then, \[\frac{{{K}_{P}}}{{{K}_{\alpha }}}={{\left( \frac{{{q}_{p}}}{{{q}_{\alpha }}} \right)}^{2}}\times \frac{{{m}_{\alpha }}}{{{m}_{p}}}\] \[\Rightarrow \] \[\frac{{{K}_{p}}}{{{K}_{\alpha }}}={{\left( \frac{{{q}_{p}}}{2{{q}_{p}}} \right)}^{2}}\times \frac{4{{m}_{p}}}{{{m}_{p}}}=1\] \[\Rightarrow \] \[{{K}_{\alpha }}=1meV\]
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