Question

If the de Broglie wavelength of a proton accelerated through a potential difference $V_1$ is same as the de Broglie wavelength of an alpha particle accelerated through $V_2$, then $V_1 : V_2 = $

• A. 1:1
• B. 8:1
• C. 1:2
• D. 4:1

Hint:

de Broglie wavelength: $\lambda = h / \sqrt{2 m q V}$.
Alpha particle: mass = 4 m_p, charge = 2 e.
Careful: include charge and mass in ratio calculations.
Voltage ratio gives matching wavelengths.

Answer 1

The Correct Option is B

• de Broglie wavelength: $\lambda = \frac{h}{p}$, $p = \sqrt{2 m q V}$.
• For proton: $\lambda_p = \frac{h}{\sqrt{2 m_p q V_1}}$, alpha: $\lambda_\alpha = \frac{h}{\sqrt{2 m_\alpha q V_2}}$.
• Equate wavelengths: $\sqrt{2 m_p V_1} = \sqrt{2 m_\alpha V_2} \implies V_1 / V_2 = m_\alpha / m_p = 4 / 1 = 4$ Wait, alpha particle has charge 2e → $V_1 / V_2 = 4 * 2 = 8$.
• Hence $V_1 : V_2 = 8:1$.