Question

If velocity of free electron is made double, change in its de-Broglie wavelength will be…….. 

(A) increase by (λ/2) 

(B) decrease by (λ/2) 

(C) increase by 2λ 

(D) decrease 2λ

Answer 1

The correct option is (B) decrease by (λ/2).

Explanation:

Given: V₂ = 2V₁ 

KE of electron = E₁ = (1/2) mV₁² ....(1)

New KE of electron E₂ = (1/2) mV₂²

E₂ = (1/2)m (2V₁)²

∴ E₂ = 4[(1/2) mV₁²]

∴ E₂ = 4E₁ ------ from (1)

also

λ = [λ / {√2mE}]

As h and m is constant,

λ ∝ {1 / √E}

 ∴ (λ₂ / λ₁) = √(E₁ / E₂) = √{E₁ / 4E₁} = (1/2)

 ∴ λ₂ = (λ₁ / 2)

 hence change in wavelength = λ₁ – λ₂

= λ₁ – (λ₁ / 2) = (λ₁ / 2)

 i.e. decreases by (λ/2)