The de-Broglie wavelength of a particle having kinetic energy E is λ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

Solution

Physics sir · 3 steps

Wavelength goes as one over the square root of energy, so square the 4/3 ratio and subtract the old E.

1. Link λ to E. Momentum and kinetic energy are related by p = √2mE. So λ = h/p = h/√2mE, which means λ ∝ 1/√E for a fixed mass.

   E'/E = (λ/λ')² = (λ/(3/4λ))² = (4/3)² = 16/9

2. New energy. With λ' = 0.75λ = 3/4λ, the new kinetic energy is E' = 16/9E.
3. So. The extra energy is E' - E = 16/9E - E = 7/9E, which is option (b).

The idea to keep. For a particle of fixed mass, λ ∝ 1/√E. Shrinking λ by a factor multiplies E by that factor squared, inverted. Always check whether the question wants total or extra energy.

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