X-ray fall on a photosenstive surface to cause photoelectric emission. Assuming that the work function of the surface can be neglected, find the relation between the de -Broglie wavelength `(lambda)` of the electrons emitted to the energy `(E_v)` of the incident photons. Draw the nature of the graph for `lambda` as a function of `E_v`.

Explanation

Since , work function of a surface can be neglected so, `phi_0=0`. From photoelectric equation
`K_(max)=E_v+phi_0=E_v and K_(max)=1/2mv_(max)^2 =((mv_(max))^2)/(2m)`
or `(mv_(max))=sqrt(2mK_(max))=sqrt(2mE_v)`
Wavelength of the emitted electron is
`lambda=h/(mv_(max))=h/(sqrt(2mE_v)), i.e., lambda prop 1/(sqrt(E_v))`.
Therefore, a graph `lambda` versus `E_v` will be a parabola as shown in Fig.