An electron and an alpha particle are accelerated by the same potential difference. Let…
An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_{\varepsilon}$ and $\lambda_a$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:
$\lambda_e\gt\lambda_\alpha$
$\lambda_e=4 \lambda_\alpha$
$\lambda_e=\lambda \alpha$
$\lambda_e \lt \lambda_a$
Solution
de-Broglie wavelength is given by
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}$
For same potential difference
$\lambda \propto \frac{1}{\sqrt{m q}}$
$\frac{\lambda_\alpha}{\lambda_e}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}}$
$\because m_\alpha \gg m_e$
$\lambda_e\gt\lambda_\alpha$