What should be the order of arrangement of de-Broglie wavelength of electron (λₑ), an α-particle (λₐ) and proton (λₚ) given that all have the same kinetic energy?

Solution

Physics sir · 4 steps

Same kinetic energy means λ ∝ 1/√m, so the lightest particle has the longest wavelength.

1. Link λ to energy. de Broglie wavelength is λ = h/p. Kinetic energy is K = p²/2m, so the momentum is p = √2mK.

   λ = h/√2mK

2. Same K. With h and K the same for all three, only the mass changes, so λ ∝ 1/√m. A heavier particle has a shorter wavelength.
3. Order the masses. The electron is about 1836 times lighter than the proton, and the α-particle is about 4 times heavier than the proton. So mₑ < mₚ < m_α.
4. So. The order of wavelengths is the reverse of the masses: λₑ > λₚ > λ_α, which is option (c).

The idea to keep. Turn λ = h/p into λ = h/√2mK whenever kinetic energy is given. At equal K, the lightest particle has the longest wavelength.

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