JEE Main 2022 · 25 July · evening · Single correct · +4 / −1
The ratio of wavelengths of proton and deuteron accelerated by potential Vₚ and V d is 1 : √2. Then the ratio of Vₚ to V d will be :
- (a) 1 : 1
- (b) √2 : 1
- (c) 2 : 1
- (d) 4 : 1
Answer and solution
Answer (d) 4 : 1
Solution
Physics sir · 4 steps
Write λ in terms of mass, charge and voltage, then let the equal charges cancel and track the masses.
- Wavelength from voltage. A charge q accelerated from rest through V gains kinetic energy qV, so its momentum is p = √2mqV. Its de Broglie wavelength is λ = h/p.
λ = h/√2mqV ⇒ λₚ/λ_d = √((m_d q_d V_d)/(mₚ qₚ Vₚ))
- Compare the particles. Proton and deuteron both carry charge e, so the charges cancel. A deuteron has one proton and one neutron, so m_d = 2mₚ. That leaves λₚ/λ_d = √(2V_d/Vₚ).
- Use the given ratio. Set √(2V_d/Vₚ) = 1/√2. Squaring gives 2V_d/Vₚ = 1/2, so Vₚ/V_d = 4.
- So. Vₚ : V_d = 4 : 1, which is option (d).
The idea to keep. For a charged particle accelerated from rest, λ ∝ 1/√mqV. Always write all three factors, then cancel only what is truly equal for the two particles.
Answer from the NTA final answer key. Solution by PrepWiser.