Question

An electron is accelerated between two points having potential 20 V and 40 V, de-Broglic wavelength of electron is……… 

(A) 0.75 Å 

(B) 7.5 Å 

(C) 2.75 Å 

(D) 0.75 nm

Answer 1

The correct option is (C) 2.75 Å.

Explanation:

We know λ = (h/P) = {h / √(2mE)} 

but E = eV_o 

Hence λ = [h / {√(2meVo)}] 

Given: potential = V_o = V₁ – V₂ 

∴ V_o = 40 – 20 = 20V 

∴ λ = [{6.62 × 10^–34} / √{2 × 9.1 × 10^–31 × 1.6 × 10–19 × 20}] 

∴ λ = 0.274 × 10^–9 m λ 

= 2.74 Å