Question

A proton and an election are accelerated by the same potential difference. Let λ_e and λ_p denote the de Broglie wavelengths of the election and the proton respectively.

1. λ_e < λ_p
2. λ_e > λ_p
3. λ_e = λp
4. The relation of λe and λp will depend on the accelerating potential difference.

Answer 1

Correct Answer - Option 2 : λ_e > λ_p

The de - Broglie wavelength (λ) is given as

\(\lambda \; = \;\frac{h}{p}\) 

Where h is the planck constant and P is the momentum of the particles.

Now, for proton P_p > P_e

⇒ λ_e > λ_p