A particle moving with kinetic energy E has de Broglie wavelength λ. If energy ΔE is added to its energy, the wavelength become λ/2. Value of ΔE, is :

Solution

Physics sir · 3 steps

Since λ ∝ 1/√E, halving the wavelength needs four times the energy, not twice.

1. Link λ and E. Momentum and kinetic energy are related by p = √2mE. Put this into λ = h/p.

   λ = h/√2mE ⇒ λ ∝ 1/√E

2. Halve the wavelength. For λ to become λ/2, √E must double. So the new energy is E' = 4E.
3. So. The energy added is Δ E = E' - E = 4E - E = 3E, which is option (b).

The idea to keep. For a particle, λ = h/√2mE, so λ goes as 1/√E. Square the wavelength ratio to get the energy ratio, then check whether they ask for the new energy or the added energy.

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