Find the photoelectric work function for this metal

[icelated]

Homework Statement

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

Homework Equations

[itex]V_0 =\frac{hf}{e}-\frac{\phi}{e}[/itex]

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?

 

[BruceW]

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?

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.

 

[technician]

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

 

[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 that's in jewels
so i convert that to ev
8.2825 / 1.6 x 10^-19 = 5.17ev

but the answer is 4.8ev

 

[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

 

[BruceW]

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

 

[icelated]

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

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