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

1. de-Broglie wavelength for a charged particle accelerated through potential $V$

$$ \lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}} $$ 

Where $h$ = Planck's constant, $m$ = mass, $q$ = charge, $V$ = accelerating potential

2. For same potential difference $V$

$$ \lambda \propto \frac{1}{\sqrt{mq}} $$

So the ratio for alpha particle $(a)$ and electron $(e)$:

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

3. Compare masses and charges

Alpha particle = He nucleus: $m_a \approx 4m_p \approx 4 \times 1836 \, m_e \approx 7344 \, m_e$

Charge: $q_a = +2e$, while $q_e = e$

$$ m_a \gg m_e \quad \text{and} \quad q_a = 2q_e $$

$$ \frac{\lambda_a}{\lambda_e} = \sqrt{\frac{m_e \cdot e}{7344 m_e \cdot 2e}} = \sqrt{\frac{1}{14688}} \approx \frac{1}{121} $$

Therefore: $\lambda_e \gg \lambda_a$ or $\lambda_e > \lambda_a$

Electron has much larger de-Broglie wavelength than alpha particle for same accelerating voltage.