The de Broglie wavelength for an electron accelerated through the potential difference of V₁ volt is λ₁. When the potential difference is changed to V₂ volt, the associated de Broglie wavelength is increased by 50%. If (V₁/V₂) = (9/α), then the value of α is ______。 Numerical answer type. Work it out, then open the answer.

Answer: 4. λ ∝ 1/√(V) ⇒ (V₁)/(V₂) = ((λ₂)/(λ₁))². λ₂ = 1.5λ₁ ⇒ (V₁)/(V₂) = 9/4 So α = 4.

The de Broglie wavelength for an electron accelerated through the… | JEE Main 2026 (6 Apr, Shift 2)

Q9JEE Main 2026 · 6 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$ . When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$ . If $(V_1/V_2) = (9/\alpha)$ , then the value of $\alpha$ is ______。

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

$\lambda \propto \dfrac{1}{\sqrt{V}} \Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2$ .
$\lambda_2 = 1.5\lambda_1 \Rightarrow \dfrac{V_1}{V_2} = \dfrac{9}{4}$
So $\alpha = 4$ .

Report a mistake

Matter Waves: de Broglie Wavelength question from JEE Main 2026 (6 Apr, Shift 2)

More questions on Matter Waves: de Broglie Wavelength