Work function and kinetic energy

[Abdul.119]

Homework Statement

The work function for potassium is 2 eV. Consider light with wave length 3.6*10^-7 m hitting the potassium surface.
a) what is the stopping potential?
b) what is the kinetic energy and velocity of the fastest electrons emitted?

Homework Equations

KE_max = hf - Φ
q_e V_s = E_photon - q_e Φ
where q_e is the charge of electron, V_s is stopping potential, and Φ is work function

The Attempt at a Solution

for finding the stopping potential, V_s = (hf/q_e) - Φ

f = c/λ , from the given wavelength, I found f = 8.33*10^14
then hf = 6.6*10^-34 * f = 5.49*10^-19
then, V_s = (5.49*10^-19 / 1.6*10^-19) - 2 eV = 1.43 volts
Correct?

And as for finding the kinetic energy, I need to use the first equation, but apparently the units of Φ should not be in eV because that wouldn't make sense, how should I calculate KE_max then?
Edit: ok I converted the hf into eV, then I got the answer 1.43 eV, which is the same as the answer for a), did I do something wrong? or is it actually suppose to be the same?

 

[ehild]

Homework Statement

The work function for potassium is 2 eV. Consider light with wave length 3.6*10^-7 m hitting the potassium surface.
a) what is the stopping potential?
b) what is the kinetic energy and velocity of the fastest electrons emitted?

Homework Equations

KE_max = hf - Φ
q_e V_s = E_photon - q_e Φ
where q_e is the charge of electron, V_s is stopping potential, and Φ is work function

The Attempt at a Solution

for finding the stopping potential, V_s = (hf/q_e) - Φ

f = c/λ , from the given wavelength, I found f = 8.33*10^14
then hf = 6.6*10^-34 * f = 5.49*10^-19
then, V_s = (5.49*10^-19 / 1.6*10^-19) - 2 eV = 1.43 volts
Correct?

Not quite.

Write out the units. What is the dimension of hf? In what units did you get it?
What is the dimension of the Work Function? Is it energy or potential?