A) 1
B) \[\sqrt{\frac{2{{m}_{e}}}{{{m}^{a}}}}\]
C) \[\sqrt{\frac{{{m}_{e}}}{{{m}_{\alpha }}}}\]
D) \[\sqrt{\frac{{{m}_{e}}}{2{{m}_{\alpha }}}}\]
Correct Answer: D
Solution :
Given: Electron and a-particle are accelerated by 100 V. Now from formula \[{{v}_{electron}}=\sqrt{\frac{2eV}{{{m}_{electron}}}}\] and \[{{v}_{\alpha }}=\frac{\sqrt{2{{e}_{\alpha }}V}}{{{m}_{\alpha }}}\] But we know that \[{{e}_{\alpha }}=2e\] Hence, \[{{\upsilon }_{\alpha }}=\sqrt{\frac{4eV}{{{m}_{\alpha }}}}\] Hence, ratio of momenta \[=\frac{{{m}_{electron}}\,{{v}_{electron}}}{{{m}_{\alpha }}{{v}_{\alpha }}}\] \[=\frac{{{m}_{electron}}}{{{m}_{\alpha }}}\times \frac{\sqrt{2eV\text{/}{{m}_{electron}}}}{\sqrt{4eV/{{m}_{\alpha }}}}\] \[=\frac{\sqrt{2eV\,{{m}_{electron}}}}{\sqrt{4eV\,{{m}_{\alpha }}}}=\sqrt{\frac{{{m}_{electron}}}{2{{m}_{\alpha }}}}\]
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