An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_e$

Question

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

Professor PC Thomas Classes · Answer

(a) : The de-Broglie wavelength in terms of potential difference is given by, \ $ \lambda=\frac{h}{\sqrt{2 m q v}} $ For same potential difference,   $ \lambda \propto \frac{1}{\sqrt{m}} 	ext { and } \lambda \propto \frac{1}{\sqrt{q}} $ So, $\lambda_e>\lambda_\alpha$ ( $\because m_\alpha$ and $q_\alpha$ is greater than $m_e$ and $q_e$ )