A) 1
B) \[\frac{{{m}_{e}}}{{{m}_{p}}}\]
C) \[\sqrt{\frac{{{m}_{p}}}{{{m}_{e}}}}\]
D) \[\sqrt{\frac{{{m}_{e}}}{{{m}_{p}}}}\]
Correct Answer: C
Solution :
If a charge particle of mass m and charge q is accelerated through a potential difference V. and E is the energy acquired by the particle, then \[E=qV\] If v is velocity of particle, then \[E=\frac{1}{2}m{{v}^{2}}\] or \[v=\sqrt{\left( \frac{2E}{m} \right)}\] Now, de-Broglie wavelength of particle is \[\lambda =\frac{h}{mv}=\frac{h}{m\sqrt{(2E/m)}}\] Substituting the value of E, we get \[\lambda =\frac{h}{\sqrt{2mqV}}\] For electron, \[{{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}\,eV}}\] For proton, \[{{\lambda }_{p}}=\frac{h}{\sqrt{2{{m}_{p}}\,eV}}\] \[\therefore \] \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\sqrt{\left( \frac{{{m}_{p}}}{{{m}_{e}}} \right)}\]
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