Answer: Maximum Work Function for 900nm IR Light

babypudding
Messages
4
Reaction score
0
Maximum work function??

Homework Statement


What is the maximum work function for a surface to emit electrons when illuminated with 900nm infrared light?


Homework Equations


K= 1/2mv2
= hf - work function

wavelength =h/p = h/mv

The Attempt at a Solution



p=h/wavelength
= 6.626 x 10-34 / 900 x 10-9
= 7.362 x 10-28 kgm/s

p=mv
v=p/m
=7.362 x 10-28 / 9.11x10-31
= 1027.66 m/s

K=1/2mv2
=1/2 x 9.11x10-31 x 1027.662
=4.81x10-25 J

frequency x wavelength = c (no sure if this is right, where c is the speed of light)
f= 3x108/ 900 x 10-9
=270

work function= hf- K
= 6.626 x 10-34 x 270 - 4.81x10-25
= -4.81J
My answer looks wrong, so I think I did it wrong. Please lead me to the right direction.
 
Physics news on Phys.org


I imagine they're looking for the workfunction to be the energy of the light at that wavelength since that will be enough to free an electron with no kinetic energy.
 


So
K=1/2mv2
=1/2 x 9.11x10-31 x 1027.662
=4.81x10-25 J

That should be right?
 


babypudding said:
So
K=1/2mv2
=1/2 x 9.11x10-31 x 1027.662
=4.81x10-25 J

That should be right?

No. The wavelength of the radiation is 900nm. This is not the wavelength of the electron. So, the value you have for the electron kinetic energy is wrong.

Read Kurdt's advice again:

I imagine they're looking for the workfunction to be the energy of the light at that wavelength since that will be enough to free an electron with no kinetic energy.

If the material had the biggest possible work function it could have and still give off an electron when illuminated with 900nm radiation, then the electrons released will have K=0. Use this kinetic energy in the photoelectric effect equation to solve for what the work function must be in this case:

hf=K-\phi
 


Ok. Shouldn't it be hf = K + work function ?
h= 6.626 x 10-34
K=0
f= 3x108 / 900 x10-9 = 3.33x1014

So hf = 0 + work function
work function = 2.21 x 10-19 J

Is this right now?
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top