What Is the Maximum Kinetic Energy of a Photoelectron with a 400 nm Wavelength?

[drawar]

Homework Statement

What is the maximum kinetic energy (in eV) of a photoelectron emitted when a surface, whose work function is 5eV, is illuminated by photons whose wavelength is 400 nm
1. -1.89
2. 1.89
3. 0
4. 3.1

Homework Equations

hc/(lambda)=work function + KE max

The Attempt at a Solution

After plugging numbers in the formula above, I get KE max=-1.89 eV, but I guess I must have done something wrong, since energy cannot be negative.

 

[wukunlin]

looks like your calculation is correct. Think about the physical significance.

 

[drawar]

thank you, I've figured it out :)
since lambda > lambda0, the photoelectric effect cannot occur, then KEmax = 0