# Homework Help: Questions on photoelectric effect

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1. May 18, 2015

### EdJr

Hello!
First of all, this is my first post here. I hope it's on the right thread.
I managed to answer most of the questions, but I think at least some of them are wrong (for example, d)). Any help would be really appreciated.

1. The problem statement, all variables and given/known data
A monochromatic light beam of wavelength λ=500nm hits a clean sodium (Na) surface. The work function of sodium is W=3.7e-19 J = 2.3 eV.
a) What is the magnitude of the momentum of one photon of this beam?
b) What are the frequency and the energy of such photon?
c) What is the cutoff frequency of sodium? Graph it.
d) Will there be emission of electrons? If so, what is the maximum kinetic energy of an emitted electron?
e) What is the cutoff potential for a cesium plate illuminated by photons of wavelength λ=400nm, considering that the maximum kinetic energy of the photoelectrons is 2.23 eV?

2. Relevant equations
p = h/λ
f = c/λ
E = hf = K+W

3. The attempt at a solution
a)
p = h/λ
p = h/500e-9
p = 1.32e-27 (kg*m)/s

b)
f = c/λ
f = c/500e-9
f = 6e14 Hz
E = hf
E ≈ 3.97e-19 J

c)
f = E/h
f = 3.7e-19/h
f ≈ 5.6e14 Hz
I think the graph is something like this, V0 being equal to W (I guess), and v0 being equal to f.

d) Since E = 3.97e-19 J is greater than W = 3.7e-19 J, there will be electron emission.
The kinetic energy will be:
E = K + W
3.97e-19 J = K + 3.7e19 J
K ≈ - 2.3 eV
I think this is wrong, because the kinetic energy shouldn't be negative, right? Is the formula wrong? Or am I using the wrong one?
(Edit: I think I just forgot the minus in the second exponent.... The result, then, should be: 3.97e-19 J = K + 3.7e-19 J ≈ 2.7e-20 J ≈ 0.1685 eV)

e)
I'm not sure how to solve this one.... Is the cutoff potential the same thing as the work function? If so, I'll run into the same problem as the previous one. If not, I have no idea how to solve this.

Last edited: May 18, 2015
2. May 18, 2015

### Staff: Mentor

Good so far (apart from the typo and missing units)
How did you get that?
Yes, kinetic energy should be positive. You have to add something positive to 3.7e-19 J to get 3.97e-19 J as the second number is larger.

Yes. (e) is like (d), just going in the opposite direction.

3. May 18, 2015

### Noctisdark

Hello, E is the energy of the incident photon, and as you said is larger that the work function, the formula you used in (d) looks fine(regardless of sign mistakes), take a look at it you just recalculate
(e)Use E=W+K
You know the kinetic energy + you know the energy of photon so work it out !!
I encourage you and sorry if I made any syntax mistake ...

4. May 18, 2015

### EdJr

Thanks for the fast responses!

Oh, that was really silly on my part. The typo was exactly what led to the wrong result. I thought the missing units weren't required at that point, but good to know.
So, it should be 3.97e-19 J = K + 3.7e-19 J, with the result being ≈ 0.1685 eV.... Right?

Hmm, ok. So, I tried this here:
f = c/λ
f ≈ 7.49e14 Hz
E = hf
E ≈ 4.966e-19 J
E = K + W
4.966e-19 J = 2.23 eV + W
W ≈ 1.39e-19 J

Is that right?
I'm still thinking that the cutoff potential and the work function aren't the same thing, since I've read in several places about them as different variables. Anyway, I couldn't find a formula to calculate it.

Thanks for the encouragement!

Last edited: May 18, 2015
5. May 18, 2015

Right.
Right.