Question

A particle of mass m and charged q is accelerated through a potential V.The Degree Broglie wavelength of the particle will be

Answer 1

This the expression.

Answer 2

A particle of mass m and charged q is accelerated through a potential V .
e.g., potential energy of particle = charge × potential difference = qV
P.E of particle = qV

now, body is accelerated , Let body moves with speed v
Then, Kinetic energy of particle = 1/2 mv²
K.E of particle = 1/2mv²
If P is the momentum of particle then, K.E of particle = P²/2m

Use conservation energy theorem,
K.E of particle = P.E of particle
qV = P²/2m
P²= 2qVm ⇒p = √{2qVm} ------(1)

Now, according to De - broglie's wavelength concept ,
λ = h/P
where λ is wavelength of particle , h is plank's constant and P is the momentum of particle .

so, λ = h/√{2qVm} from equation (1)
Hence, wavelength of particle is h/√{2qVm}