# Find the photoelectric work function for this metal

1. May 22, 2013

### icelated

1. The problem statement, all variables and given/known data

The graph in fig shows the stopping potential as a function of the frequency of the incident light falling on the metal surface.

Find the photoelectric work function for this metal

2. Relevant equations

$V_0 =\frac{hf}{e}-\frac{\phi}{e}$

3. The attempt at a solution

From the graph, fth =1.25×10 Hz.
I am completely lost.
The solution gives 4.8ev
is it h times threshold?
how do you find h?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. May 22, 2013

### BruceW

yep. the work function is just the threshold frequency times h. Although, I don't think the threshold frequency is 1.25*10 Hz... shouldn't it be 10 to the power of something? I can't tell because the graph is a bit hard to read. and how to find h? Um. It is just a constant. It should be given to you in the question, or you can look it up online.

3. May 22, 2013

### technician

h is the gradient of the straight line, you need 'e' to convert stopping potential into energy in joules

4. May 22, 2013

### icelated

The solutions said fth =1.25×10^15 Hz.
h = planks constant = 6.26x10^-36
if you multiply you get 8.2825 i have to assume thats in jewels
so i convert that to ev
8.2825 / 1.6 x 10^-19 = 5.17ev

5. May 22, 2013

### icelated

I think i got it.
h= (e)(slope) = (1.60×10−16 C)(3.8×10−15 V ⋅ s) = 6.1×10^−34 J ⋅ s
(6.1×10^−34 Js)(1.25x10^15) = 7.75x10^-19
then convert to ev
7.75x10^-19 / 1.6x10^-19 = 4.84 x10^-19

6. May 22, 2013

### BruceW

ah, good work. I see, you got h from the graph. nice, I didn't think of that.

7. May 22, 2013

### icelated

Yes, i got h from the graph. Thank you.