Question

Solve this (Chap : Atomic Structure)

Answer 1

☣️☣️☣️For each electron of mass M and charge q  inside a potential difference V,  just before collision with a target atom, we can say that its P.E. and K.E. equal.☣️☣️☣️

☣️☣️P.E. = K.E.

☣️☣️qV = (1/2)Mv2

☣️☣️2MqV = M2v2  or

☣️☣️ Mv = 2MqV ½

☣️☣️ λ = h/Mv
λ = h / √ 2MqV

⚛️⚛️Direct formula
⚛️⚛️λ = (150/V)½ A°

Answer 2

Answer:

see the attachment......