Threshold wavelength for photo-electric emission of a metallic surface is 3800Armstrong.A beam of light of wavelength 2600armstrong is incident on the surface.what will be the work function of the metal?

1 Answer
Jan 21, 2018

2.41 xx10^-19J or 1.51 eV

Explanation:

E = hf
where h is Planck's constant, f is the frequency of photo.

c = flamda

rArrE = (hc)/lamda

Let

E_1 be the incident photo energy
lamda_1 be the wavelength of the incident photo#

E_2 be the energy of the emitted photo.
lamda_2 wavelength of the emitted photo.

Obviously E_1 has higher energy because the incident photo has shorter wavelength then the outgoing photo.

It begs the question: where does rest of the energy go?
It goes to overcome the work function of the metal so that the emitted photo can escape the metal.

Work function = E_1-E_2 = (hc)/lamda_1- (hc)/lamda_2 =(lamda_2-lamda_1)/(lamda_1*lamda_2)hc

= (380 cancel(nm)-260cancel(nm))/(380cancel(nm)260ncancelm)*( 6.626xx10^(-34) Jcancels*3xx10^8cancelm/cancels) = 2.41 xx10^-19J

Work function = 2.41 xxcancel(10^-19J) xx (1.0 eV)/(1.602 xx cancel(10^-19J)) = 1.51 eV