Question

An electron and an alpha particle are accelerated by the same potential difference. Let λ_e and λ_α 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

The Correct Option is D

The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$. When the potential difference (V) is constant, the wavelength is inversely proportional to the square root of the product of mass (m) and charge (q), i.e., $\lambda \propto \frac{1}{\sqrt{mq}}$.

Therefore, the ratio of wavelengths is $\frac{\lambda_a}{\lambda_e} = \sqrt{\frac{m_e q_e}{m_a q_a}}$.

Given that $m_a >> m_e$ and $q_a = 2q_e$, it follows that $\lambda_e > \lambda_a$.