Determination of planck's constant

[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?

 

[robphy]

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

 

[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.

 

[rock.freak667]

the work done in stopping the electron is equal to the kinetic energy lost...
that is,
[itex]eV_s=\frac{1}{2}mv^2[/tex]<br /> <br /> so from $hf=hf_0+\frac{1}{2}mv^2$=>$hf=hf_0+eV_s$<br /> <br /> rearrange and make it into the form you want...[/itex]

 

[Ry122]

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

 

[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?

 

[Ry122]

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

 

[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

 

[Ry122]

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

 

[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.