Question

Relation between kinetic energy and wavelength of photoelectron

Answer 1

*A2A

⟹K.E=p22m⟹K.E=p22m

⟹p=2m(K.E)−−−−−−−−√⟹p=2m(K.E)

where,

K.EK.E = Kinetic Energy

mm = Mass of object

pp = Momentum of object

⋆⋆ The wavelength of the object is given by its de-Broglie wavelength(λ)(λ) :-

⟹λ=hp⟹λ=hp

⟹λ=h2m(K.E)−−−−−−−−√⟹λ=h2m(K.E)

where,

hh = Planck’s constant