JEE Main 2023 · 11 April · evening · Single correct · +4 / −1
The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :
(Assume mₚ=mₑ × 1849 )
- (a) 1:43
- (b) 1:62
- (c) 2:43
- (d) 1:30
Answer and solution
Answer (a) 1:43
Solution
Physics sir · 4 steps
Write λ in terms of kinetic energy. With K the same, only the mass is left, under a square root.
- Momentum from energy. Kinetic energy is K = p²/2m, so the momentum is p = √2mK. Put this into λ = h/p.
λ = h/√2mK
- Take the ratio. Both particles have the same K, so h, the 2 and K cancel. That leaves λₚ/λₑ = √(mₑ/mₚ).
- Put in numbers. With mₚ = 1849 mₑ, the ratio is √(1/1849) = 1/43, because 43² = 1849.
- So. λₚ : λₑ = 1 : 43, which is option (a). The heavier proton has the shorter wavelength.
The idea to keep. For the same kinetic energy, λ ∝ 1/√m. For the same speed, λ ∝ 1/m. Always check which quantity is held equal before you take the ratio.
Answer from the NTA final answer key. Solution by PrepWiser.