Photoelectrons and Planck's constant

[benca]

Homework Statement

Use the graph to determine Planck's constant.

Relevant Equations

E = hf
Ek = hf - W

h = Planck's constant
f = frequency
W = work function

Attempt:

I was thinking of finding the slope of the graph but I only know the values for x = 10, y = 3 and y = 0. And without the y-intercept, I don't know the work function and can't solve for h. If you can't see from the picture, the last co-ordinate is (10,3) and the x-axis is measured in f x 10^14 Hz

I'm not sure what options I have left if I don't know how to figure out the slope or work function.

 

[gneill]

The graph is of a straight line. Could you not simply extrapolate the line to find the y-intercept?

 

[benca]

I thought of that too, but I wouldn't exactly get Planck's constant but instead a number close to Planck's constant. But if that's what I need to do then that's ok, thanks

 

[gneill]

I thought of that too, but I wouldn't exactly get Planck's constant but instead a number close to Planck's constant. But if that's what I need to do then that's ok, thanks

Yup. You need to squeeze what you can from the given data. That's a truism for experimental science, where you collect measured data over a limited region and suggest reasonable extrapolations to complete the overall picture for purposes of analysis. In this case the graph is clearly linear, so a linear extrapolation would not be an unreasonable proposition.

 

[benca]

Yup. You need to squeeze what you can from the given data. That's a truism for experimental science, where you collect measured data over a limited region and suggest reasonable extrapolations to complete the overall picture for purposes of analysis. In this case the graph is clearly linear, so a linear extrapolation would not be an unreasonable proposition.

Alright, thanks

 

[gneill]

Alright, thanks

Happy to be of help! Cheers!

 

[Charles Link]

The straight line has the form $y=mx+b$. In this case $y=E_k$ and $x=f$. The slope is found as Planck's constant $h=m=\frac{y_2-y_1}{x_2-x_1}$. From what I could see, you apparently need a brush up on your algebra. Here $(x_1,y_1)$ and $(x_2,y_2)$ are any two points on the straight line. $\\$ In addition, once you have $m$, you can then write $m=\frac{y-y_1}{x-x_1}$, and ultimately solve for $b=-W$, which is the y-intercept. This problem is a simple one, but you need to be able to do algebra.

 

[Orodruin]

I thought of that too, but I wouldn't exactly get Planck's constant but instead a number close to Planck's constant. But if that's what I need to do then that's ok, thanks

You never get an exact value from measurements. That is the nature of empirical sciences.

Also, to spell out what others have noted: You only need two points on a line to determine its coefficients. Three will give you an overconstrained system and with measurement errors present you will typically need something like a least squares method to determine the best fitting line.

 

[Charles Link]

On this one, if you pick $(10,3)$ and $(2.75,0)$ as your two points, you do get close to two decimal place accuracy. $\\$ One thing that should be mentioned is that energy is in eV here, and you need to convert to joules to get a number that is approximately the accepted value of Planck's constant $h=6.626 E-34$ joule-sec. Otherwise, you get the number in eV-seconds. I would venture to guess that most physics people know this number as 6.626E-34 joule-sec, but don't know its value in units of eV-seconds.

 

[Orodruin]

I would venture to guess that most physics people know this number as 6.626E-34 joule-sec

Most physicists I know (admittedly mainly particle phycisists) know the number as $2\pi$ due to preferring natural units.

 

[Charles Link]

Most physicists I know (admittedly mainly particle phycisists) know the number as $2\pi$ due to preferring natural units.

That is a much more advanced version, where $\hbar=c=1$, but here we are at the Physics 101 level.

 

[Orodruin]

That is a much more advanced version, where $\hbar=c=1$, but here we are at the Physics 101 level.

In a few weeks I will be giving an ”inspirational” lecture about the geometry of relativity to the same students I taught vector analysis the past spring and went on and on about making sure their physical dimensions always work out ... letting c = 1 should be great fun!

 

[Charles Link]

Hopefully the OP @benca returns to complete the exercise. If he works it through, I think he will be surprised how close the result he gets is to $h=6.626E-34$. From the data that was supplied, you can't get 3 decimal place accuracy, but you can get pretty close to two decimal places.

 

[DaveE]

That is a much more advanced version, where $\hbar=c=1$, but here we are at the Physics 101 level.

As you suggested, much too advanced for an introductory physics homework forum.