Question

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?

Answer 1

`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`