Answer: Maximum Work Function for 900nm IR Light

[babypudding]

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.

 

[Kurdt]

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.

 

[babypudding]

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

That should be right?

 

[G01]

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:

$$hf=K-\phi$$

 

[babypudding]

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?