JEE Main 2026 Physics question
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 .
Answer: 4
Solution
λ∝1/√V, so V₁/V₂=(λ₂/λ₁)²=(1.5)²=9/4. Hence α=4.