Question

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

Answer 1

According to De- Broglie's theory,
any subatomic particle have both particle as well as wave nature .
wavelength can be calculated by the ratio of Plank's constant to momentum of subatomic particle.
e.g., λ = h/p

here , A particle of mass m and charged q is accelerated through a potential V .
so, potential energy of particle = qV
Let body moves with velocity = v
then, kinetic energy of particle = 1/2 mv²

according to conservation of energy theorem,
now, kinetic energy of particle = potential energy of particle
1/2mv² = qV
we know, momentum , P = mv
so, 1/2mv² = P²/2m use it above,

P²/2m = qV
P = √{2qVm}

now, wavelength of particle = h/p
λ = h/√{2qVm}