JEE Main 2026 · 6 Apr · Shift 2 · Physics Q47

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PhysicsInteger+4 / -1Question 47 of 75

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 ________.

Your answer

Answer and worked solution

Official answer: 4

de Broglie. $\lambda\propto\dfrac1{\sqrt V}$ . $\lambda_2=1.5\lambda_1$ gives
$\frac{V_1}{V_2}=2.25=\frac94.$

Answer: 4

Topics:de Broglie waves

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JEE Main 2026 · 6 Apr · Shift 2, Physics question 47