Question

TItanium(6.94x10-190 and silicon (7.77x10-19 J) surfaces are irradiated with UV radiation with a wavelength of 251 nm. what is the wavelength of the electrons emitted by the titanium surface?

Answer 1

First we will find energy of incident photon

E = h = hc/

h = planck's constant = 6.63 x 10^-34 Js

c = speed of light = 3 x 10⁸ m/s

= wavelength of incident photon = 251 nm = 2.51 x 10^-7 m.

E =  6.63 x 10^-34 x 3 x 10⁸/ 2.51 x 10^-7 = 19.89 x 10^-26 / 2.51 x 10^-7 = 7.92 x 10^-19 J

Now the energy of the emitted electron:

Energy of emitted electron, E(k) = Energy of incident photon (E) - work function ()

(Work function of any solid surface is the minimum energy required to remove an electron to infinity from the surface )

Work function for Titanium = 6.94x10^-19 J

E(k) = 7.92 x 10^-19 J - 6.94x10^-19 J = 0.98 x 10^-19 J

Now, E(k) = hc/

=  hc/E(k) =  6.63 x 10^-34 x 3 x 10⁸/ 0.98 x 10^-19 =  19.89 x 10^-26 / 0.98 x 10^-19 = 20.29 x 10^-7 m = 2029 nm