The de Broglie wavelengths for an electron and a photon are λₑ and λₚ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two ?

Solution

Physics sir · 4 steps

Write the same kinetic energy K in terms of each wavelength, then set the two expressions equal.

1. Electron. For a particle with mass, p = √2mK, so λₑ = h/√2mK. Squaring and rearranging gives K = h²/2mλₑ².
2. Photon. A photon has no rest mass. Its whole energy is K = hc/λₚ, so λₚ = hc/K.
3. Same energy. Put the electron's expression for K into the photon's wavelength.

   λₚ = hc/K = hc × 2mλₑ²/h² = 2mc/h λₑ²

4. So. 2mc/h is a constant, so λₚ ∝ λₑ². That is option (a).

The idea to keep. For a massive particle λ ∝ 1/√K, but for a photon λ ∝ 1/K. Whenever you compare the two at equal energy, the photon's wavelength goes as the square of the electron's.

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