A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume h = 6.63 × 10⁻³⁴ J s, mₑ=9.0 × 10⁻³¹ kg and mₚ=1836 times mₑ )

Solution

Physics sir · 3 steps

Same wavelength means same momentum, so kinetic energy depends only on which particle is lighter.

1. Same momentum. The de Broglie wavelength is λ = h/p. If the proton and the electron have the same λ, they must have the same momentum p.

   K = p²/2m ⇒ Kₚ/Kₑ = mₑ/mₚ

2. Put in masses. With p the same for both, K ∝ 1/m. Since mₚ = 1836 mₑ, we get Kₚ/Kₑ = mₑ/(1836 mₑ) = 1/1836. The values of h and mₑ are not needed.
3. So. Kₚ : Kₑ = 1 : 1836. The heavier proton carries less kinetic energy for the same momentum, which is option (c).

The idea to keep. Equal de Broglie wavelengths always mean equal momenta. Then use K = p²/2m: kinetic energy goes as 1/m, so the lighter particle has more energy.

Answer from the NTA final answer key. Solution by PrepWiser.