The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25 % of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

Solution

Physics sir · 3 steps

Same wavelength means same momentum, so compare pv/2 for the electron with pc for the photon.

1. Same momentum. de Broglie says λ = h/p for any particle, photon included. Equal wavelengths therefore give the electron and the photon the same momentum p.
2. Two energy formulas. For the electron, Kₑ = p²/2m = pv/2, using p = mv. For the photon, E = pc.

   Kₑ/Eₚₕₒₜₒₙ = (pv/2)/pc = v/2c

3. So. With v = 0.25 c, the ratio is 0.25/2 = 1/8, which is option (c).

The idea to keep. Equal de Broglie wavelengths mean equal momenta. Then write each energy in terms of p: p²/2m for a massive particle and pc for a photon.

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