Question

the de broglie wavelength associated with an electron accelerated through a potential difference V is lambda. what will be the de broglie wavelength when the accelerating potential difference is increased to 4 volt?​

Answer 1

Answer:

Explanation:

CASE - 1

WAVELENGTH = λ₁

POTENTIAL DIFFERENCE = V₁

CASE - 2

WAVELENGTH = λ₂

POTENTIAL DIFFERENCE = V₂ = V₁

NOTE :

λ∝ 1/√V

λ₁/λ₂ = √V₂/√V₁

       = √4/1

       = 2

λ₁ = 2 λ₂.

HOPE THIS HELPS.