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  4. An electron of mass m, when accelerated through a potential difference V, has de Broglie wavelength λ. The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be

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Q. An electron of mass $m$, when accelerated through a potential difference $V$, has de Broglie wavelength $\lambda$. The de Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference, will be

AIPMTAIPMT 1995Dual Nature of Radiation and Matter

Solution:

Momentum of electrons $\left(p_{ e }\right)=\sqrt{2 m e V}$
and momentum for proton $\left(p_{p}\right)=\sqrt{2 M e V}$
Therefore, $\frac{\lambda_{ p }}{\lambda_{ e }}=\frac{h / p_{p}}{h / p_{e}}=\frac{p_{e}}{p_{p}}=\frac{\sqrt{2 m e V}}{\sqrt{2 M e V}}=\sqrt{\left(\frac{m}{M}\right)}$.
Therefore $\lambda_{p}=\lambda \sqrt{\left(\frac{m}{M}\right)}$.