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An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_\alpha$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:
Detailed Solution
De-Broglie wavelength $\lambda=\dfrac{h}{\sqrt{2mqV}}$, so $\lambda\propto\dfrac{1}{\sqrt{mq}}$. Since $m_\alpha\gg m_e$, $\lambda_e>\lambda_\alpha$.