# Maximum work function?

1. Jul 22, 2008

### babypudding

Maximum work function??

1. The problem statement, all variables and given/known data
What is the maximum work function for a surface to emit electrons when illuminated with 900nm infrared light?

2. Relevant equations
K= 1/2mv2
= hf - work function

wavelength =h/p = h/mv

3. 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.

2. Jul 22, 2008

### Kurdt

Staff Emeritus
Re: Maximum work function??

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.

3. Jul 23, 2008

### babypudding

Re: Maximum work function??

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

That should be right?

4. Jul 23, 2008

### G01

Re: Maximum work function??

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.

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:

$$hf=K-\phi$$

5. Jul 24, 2008

### babypudding

Re: Maximum work function??

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?