# Planck's Constant and the Work Function

1. Aug 11, 2009

### Procrastinate

I have recently graphed my Planck's Constant data from my Planck's Constant kit on Excel. The R squared value is great but there are a few things that confuse me.

1. I presume that the Equation of V = f(h/e) - W/e can be used with y = mx + c

If so, then I presume that my equation of y = 3e-15x - 1.2163 from excel can solve the equation:

c = -1.2163 = The work function (I find this a bit strange though because the cathode is covered in cesium which is meant to be 2.13 eV)

V = y = backing voltage

Is this right? Or do are my presumptions wrong?

Also is the letter for frequency f or v? Because there seems to be some variation whenever I attempt to browse sites.

Last edited: Aug 11, 2009
2. Aug 11, 2009

### Redbelly98

Staff Emeritus
That's the work function for pure cesium. Your cathode could have some cesium oxide or other impurity that lowers the work function.

Note quite. This should be
f(h/e) = mx
The variable in f(h/e) would equate to x, and the constants would equate to the slope m.

Both are used, though it's actually the Greek letter nu, not v.
f is a more general or universal symbol, referring to the frequency of anything that oscillates sinusoidally. ν is used strictly for the frequency of electromagnetic waves -- but the more general f can be used for those as well.

3. Aug 11, 2009

### Procrastinate

I also calculated my percentage error to be 25.4%. Is this alright or should I consider re-doing it again?

4. Aug 12, 2009

### bm0p700f

How did you calculate that.

5. Aug 12, 2009

### Procrastinate

To calculate the experimental Planck’s constant:

V = f(h/e) – W

=3.132×〖10〗^(-15) x - 1.22
This correlates to the equation
y=mx+c

∴m=3.132×〖10〗^(-15) and ∴c=- 1.22

According to theory, the gradient = h/e

∴h/(1.6×〖10〗^(-19) )=3.132×〖10〗^(-15)
∴h=3.132×〖10〗^(-15)×1.600×〖10〗^(-19)
∴h=5.011×〖10〗^(-34) J

Experimental Uncertainty:
Percentage Error=(| Experimental Value-Accepted Value |)/(Accepted Value)×100%
=(| 5.011×〖10〗^(-34)-6.626×〖10〗^(-34) |)/(6.626×〖10〗^(-34) )×100%
=0.244×100%
=24.4%

6. Aug 12, 2009

### Redbelly98

Staff Emeritus

I do wonder how many significant figures are appropriate for your 5.011e-34 value. Did you make the best-fit line by hand, or did you use a calculator or computer to get it?

7. Aug 13, 2009

### Procrastinate

I used the computer to get the R squared value and the equation. Microsoft Excel 2007.

8. Aug 13, 2009

### Redbelly98

Staff Emeritus
Looks good then. By the way, errors are usually reported to no more than 2 significant figures.