A) \[\sqrt{8}\]
B) \[\frac{1}{\sqrt{8}}\]
C) \[1\]
D) \[2\]
Correct Answer: A
Solution :
: When a charged particle of charge q and mass m is accelerated under a potential difference V, let \[\upsilon \] be velocity acquired by the particle. Then \[qV=\frac{1}{2}m{{\upsilon }^{2}}\] or \[m\upsilon =\sqrt{2mqV}\] de Broslie wavelength, \[\lambda =\frac{h}{m\upsilon }=\frac{h}{\sqrt{2mqV}}\] \[\lambda \propto \frac{1}{\sqrt{mq}}\] for the same value of V. \[\therefore \] \[\frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\frac{\sqrt{{{m}_{\alpha }}{{q}_{\alpha }}}}{\sqrt{{{m}_{p}}{{q}_{p}}}}=\sqrt{\frac{4m}{m}\times \frac{2e}{e}}=\sqrt{8}\]
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