Question

An electron and an alpha particle are accelerated by the same potential difference. Let λe and λα denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

• A. $\lambda_e > \lambda_a$
• B. $\lambda_e = 4\lambda_a$
• C. $\lambda_e = \lambda_a$
• D. $\lambda_e < \lambda_a$

Answer 1

Answer: (D)

$\lambda_e < \lambda_a$

Answer 2

de-Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$ For the same potential difference (V), $\lambda \propto \frac{1}{\sqrt{mq}}$.

$\frac{\lambda_a}{\lambda_e} = \sqrt{\frac{m_e q_e}{m_a q_a}}$

Since $m_a >> m_e$ and $q_a = 2q_e$, $\lambda_e > \lambda_a$.