Given:
Kinetic energy(K.E.)=4.55×10−29J
Mass of electron(me)=9.1×10−31kgh=6.6×10−34J−sec
debroglie wavelength λd=mvh=2mKEh
Here λ is the de Broglie wavelength, h is Planck's constant, mis the mass of the particle, v is its velocity, and p is the momentum of the particle, which is equal to mv.
λ=2mKEh
=2×9×10−31×4.5×10−296.6×10−34=9×10−306.6×10−34=96.6×10−4
=966×10−5
=7.33×10−5