An electron, a doubly ionized helium ion (He⁺⁺) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths λₑ, λ He⁺⁺ and λₚ is :

Solution

Physics sir · 4 steps

Same kinetic energy means λ depends only on mass: the lighter the particle, the longer its wavelength.

1. Link λ to energy. Kinetic energy and momentum are related by K = p²/2m, so p = √2mK. Put this into λ = h/p.

   λ = h/√2mK ⇒ λ ∝ 1/√m (same K)

2. Compare masses. The charge does not appear anywhere, because the kinetic energy is already given. Only mass matters: mₑ ≈ 1/1836 u, mₚ ≈ 1 u and m_(He²⁺) ≈ 4 u (2 protons and 2 neutrons).
3. Order the wavelengths. Lightest mass gives the longest wavelength, so mₑ < mₚ < m_(He²⁺) gives λₑ > λₚ > λ_(He²⁺). In fact λₚ = 2λ_(He²⁺), since the helium ion is 4 times heavier.
4. So. The correct relation is λₑ > λₚ > λ_(He²⁺), which is option (c).

The idea to keep. For equal kinetic energy, λ = h/√2mK, so only mass decides it. Charge enters only when particles are accelerated through the same voltage, since then K = qV.

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