# Determination of planck's constant

1. Dec 3, 2007

### Ry122

I have a graph of voltage verse frequency.
Given that the charge of an electron = 1.6 x 10^-19 I need to obtain an estimate of the value planck's constant. What equation do I need to use? Do I find the gradient of the graph?

2. Dec 3, 2007

### robphy

What is the experiment that led to these measurements (this graph)?
[I have a guess... but you should really describe it.]

3. Dec 3, 2007

### Ry122

its the stopping voltage versus frequency for a metal surface.
The only part of the graph shown is from 5.4 to 6.8 f(X 10^14) and .26 to .85 volts. The graph is linear.

4. Dec 3, 2007

### rock.freak667

the work done in stopping the electron is equal to the kinetic energy lost...
that is,
$eV_s=\frac{1}{2}mv^2[/tex] so from [itex]hf=hf_0+\frac{1}{2}mv^2$=>$hf=hf_0+eV_s$

rearrange and make it into the form you want...

5. Dec 3, 2007

### Ry122

How can I solve for h when h is on both sides of the equation?

6. Dec 3, 2007

### rock.freak667

$eV_s=hf-hf_0$
$V_s=\frac{h}{e}f-\frac{h}{e}f_0$

Y=MX+C, can you see what M and C are?

7. Dec 3, 2007

### Ry122

I can't see what they are.
The values are
m = .421428
c = -2.2757 x 10^14

8. Dec 4, 2007

### rock.freak667

Well from $V_s=\frac{h}{e}f-\frac{h}{e}f_0$

if a graph of $V_s$ and f is drawn, then the gradient,m. would be equal to the fraction: $\frac{h}{e}$ and so you know what e is, find h

9. Dec 4, 2007

### Ry122

I tryed this but but solving for h in m=h/e does not give planck's constant.

10. Dec 4, 2007

### tyco05

Are you being careful with your units when you are calculating your gradient????

Remember that your frequency scale will probably be (x 10^14) Hz

You should get a much smaller gradient.