A) \[{{\left( \frac{{{m}_{p}}}{{{m}_{e}}} \right)}^{1/2}}\]
B) \[\frac{{{m}_{e}}}{{{m}_{p}}}\]
C) \[\frac{{{m}_{p}}}{{{m}_{e}}}\]
D) 1
Correct Answer: A
Solution :
de-Broglie wavelength\[(\lambda )\] is given by \[\lambda =\frac{h}{p}\] where h is Planck's constant, and p the momentum. Also, \[p=\sqrt{2mE}=\sqrt{2mqV}\] where E is energy, q the charge and V the potential difference. \[\therefore \] \[\lambda =\frac{h}{\sqrt{2mqV}}\] Hence, the ratio of wavelength of electron\[({{\lambda }_{e}})\]and proton\[({{\lambda }_{P}})\]is \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{e}}}}={{\left( \frac{{{m}_{p}}}{{{m}_{e}}} \right)}^{1/2}}\]
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