Calculate work, frequency and Plank's constant

[Jeff97]

Homework Statement

From the graph, find the value, including its SI unit, of the following:
i. the threshold frequency;
ii. the work function of the surface;
iii. Planck's constant

Relevant Equations

N/A

i. the threshold frequency;
3.9x10^14hz? it appears the line intersects at 3.9ii. the work function of the surface;
6.626x10^-34x3.9x10^14= 2.58414 × 10^-19J

iii. Planck's constant
unsure

 

[collinsmark]

Homework Statement:: From the graph, find the value, including its SI unit, of the following:
i. the threshold frequency;
ii. the work function of the surface;
iii. Planck's constant
Relevant Equations:: N/A

View attachment 259361
i. the threshold frequency;
3.9x10^14hz? it appears the line intersects at 3.9ii. the work function of the surface;
6.626x10^-34x3.9x10^14= 2.58414 × 10^-19J

iii. Planck's constant
unsure

I'm assuming that the graph involves the photoelectric effect.

Parts i. and ii. look good to me.

For part iii., there are a few different possible approaches, but you should probably use one that doesn't rely on already knowing Planck's constant at the beginning.

What is the relationship between $\Delta E$ and $\Delta f$? In other words, what is the change in energy per unit change in frequency?

 

[Jeff97]

Yes the graph does invole the photoelctric effect.

What is the relationship between $\Delta E$ and $\Delta f$? In other words, what is the change in energy per unit change in frequency?

h = E/f so 2.58414 × 10^-19J/3.9x10^14hz = 6.626×10^-34J

 

[collinsmark]

Yes the graph does invole the photoelctric effect.

h = E/f so 2.58414 × 10^-19J/3.9x10^14hz = 6.626×10^-34J

Well, technically yes, that's one approach, sort of.

But I don't really like that approach because your $2.58414 \times 10^{-19} \ \mathrm{J}$ figure relied on Planck's constant as part of the calculation.

Wouldn't it be nice to determine Planck's constant straight from the graph?

Again, what is the change in energy per unit change in frequency? (i.e., what is $\frac{\Delta E}{\Delta f}$?)

 

[Jeff97]

Well, technically yes, that's one approach, sort of.

But I don't really like that approach because your $2.58414 \times 10^{-19} \ \mathrm{J}$ figure relied on Planck's constant as part of the calculation.

Wouldn't it be nice to determine Planck's constant straight from the graph?

Again, what is the change in energy per unit change in frequency? (i.e., what is $\frac{\Delta E}{\Delta f}$?)

Yes as I posted this I did think my approach was right. Is this the rise over run. If so I'm finding it a bit hard to
get the run I was trying this (rise×1.6×10^−19)/(runx10^14) but its wans't getting me a result even near planks constant

 

[etotheipi]

Yes as I posted this I did think my approach was right. Is this the rise over run. If so I'm finding it a bit hard to
get the run I was trying this (rise×1.6×10^−19)/(runx10^14) but its wans't getting me a result even near planks constant

I think it should be rise x 10^-19, as opposed to rise x 1.6 x 10^-19.

But the method is right, if $E_k = hf - \phi$ then $\frac{dE_k}{df} = h$, as @collinsmark mentioned.

 

[collinsmark]

Yes as I posted this I did think my approach was right. Is this the rise over run. If so I'm finding it a bit hard to
get the run I was trying this (rise×1.6×10^−19)/(runx10^14) but its wans't getting me a result even near planks constant

It might help if you actually draw out a right triangle and label the rise and run.

Your values of rise and run ($1.6 \times 10^{-19}$ and $10^{14}$) are not consistent with each other.

Again, it might help to draw out a triangle first.

 

[Jeff97]

It might help if you actually draw out a right triangle and label the rise and run.

Your values of rise and run ($1.6 \times 10^{-19}$ and $10^{14}$) are not consistent with each other.

Again, it might help to draw out a triangle first.

I have drawn a triangle.

I tried using point (8,2.6) and (3.9,0) e.g (2.6×10^−19)/(4.1x10^14) but that got me 6.34× 10^-34J close but not planks constant.

 

[etotheipi]

I have drawn a triangle.

I tried using point (8,2.6) and (3.9,0) e.g (2.6×10^−19)/(4.1x10^14) but that got me 6.34× 10^-34J close but not planks constant.

That's only about 4% off, which appears to be an acceptable degree of error to me at least. I'd be surprised if the data gave you a perfect value!

In addition to errors from reading from the graph itself, e.g. due to the width of the trendline.

 

[Jeff97]

It might help if you actually draw out a right triangle and label the rise and run.

Your values of rise and run ($1.6 \times 10^{-19}$ and $10^{14}$) are not consistent with each other.

Again, it might help to draw out a triangle first.

what do you think collin?

 

[collinsmark]

I have drawn a triangle.

I tried using point (8,2.6) and (3.9,0) e.g (2.6×10^−19)/(4.1x10^14) but that got me 6.34× 10^-34J close but not planks constant.

Your answer is going to be an approximation, no matter what you do. It's only going to be as precise as we can roughly measure from the graph.

But if you're trying to minimize your sources of error, don't use the point ($3.9 \times 10^{-19}$ J, $0$ Hz), since that point relies on your previous extrapolation attempt (and could also be a little more precise if you were more careful).

Try picking two points that are already on the given line; preferably two points that far away from each other. I chose
($8 \times 10^{14}$ J, $2.6 \times 10^{-19}$ Hz)
($5 \times 10^{14}$ J, $0.7 \times 10^{-19}$ Hz)

[Edit: Oops. I just realized that I had my units swapped. Correction below:

($8 \times 10^{14}$ Hz, $2.6 \times 10^{-19}$ J)
($5 \times 10^{14}$ Hz, $0.7 \times 10^{-19}$ J)]

The result from that is still an approximation, but it's closer to the truth.

[Edit: also, the experimentalist who created the data for the graph might have made some slight errors himself/herself. For this problem, I'm assuming that you should just trust the data; i.e., your final answer might be a little bit in error if the given data is a little bit in error. That's the best you can do.]

 

[Jeff97]

Your answer is going to be an approximation, no matter what you do. It's only going to be as precise as we can roughly measure from the graph.

But if you're trying to minimize your sources of error, don't use the point ($3.9 \times 10^{-19}$ J, $0$ Hz), since that point relies on your previous extrapolation attempt (and could also be a little more precise if you were more careful).

Try picking two points that are already on the given line; preferably two points that far away from each other. I chose
($8 \times 10^{14}$ J, $2.6 \times 10^{-19}$ Hz)
($5 \times 10^{14}$ J, $0.7 \times 10^{-19}$ Hz)

The result from that is still an approximation, but it's closer to the truth.

[Edit: also, the experimentalist who created the data for the graph might have made some slight errors himself/herself. For this problem, I'm assuming that you should just trust the data; i.e., your final answer might be a little bit in error if the given data is a little bit in error. That's the best you can do.]

Wait so how did you calculate yours?

 

[etotheipi]

@Jeff97 We've all done the same calculation. It's just every person is going to use a slightly different set of points and will get a slightly different answer. In an exam question there would be an allowed tolerance. But as @collinsmark said, it's good practice to use the largest possible triangle and avoid further extrapolation.

If you want a more accurate value, you could put the data into excel and try generating a least squares regression line.

 

[Jeff97]

@Jeff97 We've all done the same calculation. It's just every person is going to use a slightly different set of points and will get a slightly different answer. In an exam question there would be an allowed tolerance. But as @collinsmark said, it's good practice to use the largest possible triangle and avoid further extrapolation.

If you want a more accurate value, you could put the data into excel and try generating a least squares regression line.

From collins points (8,2.6) and (5,0.7) isn't the rise 3 and the run 1.9?
(3×10^−19)/(1.9x10^14) 1.58 × 10^-33J ? ?

 

[collinsmark]

From collins points (8,2.6) and (5,0.7) isn't the rise 3 and the run 1.9?
(3×10^−19)/(1.9x10^14) 1.58 × 10^-33J ? ?

Sorry for some confusion, but I realize now that I had my units swapped in my previous post. I made the correction in that post. The points I meant should be:

($8 \times 10^{14}$ Hz, $2.6 \times 10^{-19}$ J)
($5 \times 10^{14}$ Hz, $0.7 \times 10^{-19}$ J)

So the slope would then be: $\frac{2.6 \times 10^{-19} - 0.7 \times 10^{-19} \ [\mathrm{J}]}{8 \times 10^{14} - 5 \times 10^{14} \ [\mathrm{Hz}]}$

 

[Jeff97]

Sorry for some confusion, but I realize now that I had my units swapped in my previous post. I made the correction in that post. The points I meant should be:

($8 \times 10^{14}$ Hz, $2.6 \times 10^{-19}$ J)
($5 \times 10^{14}$ Hz, $0.7 \times 10^{-19}$ J)

So the slope would then be: $\frac{2.6 \times 10^{-19} - 0.7 \times 10^{-19} \ [\mathrm{J}]}{8 \times 10^{14} - 5 \times 10^{14} \ [\mathrm{Hz}]}$

Right! so doing your calculation I get 6.33 × 10^-34J

 

[collinsmark]

Right! so either doing your calculation or finding the rise and run I get 6.33 × 10^-34

Correct (Don't forget your units).

It's not precisely Planck's constant, but that's about the best you can do with the data you were given.

 

[collinsmark]

One more thing,

When you calculated the "Work function" of the surface, you used a previously known value for Planck's constant.

There's another way to do that directly from the plot (if you're not supposed to use a known Planck's constant value).

It involves finding the y-intercept.

 

[collinsmark]

The y-intercept is -2.4667 so basically -2.5 so what formula/equation do I use?

Don't forget your units and exponent.

I'm just saying that -- ignoring a minus sign -- the y-intercept is the work function at the surface.

 

[Jeff97]

oh so the 2.5 is the work function i.e 2.5x10^-19J

Also @collinsmark what would happen if the metal is replaced with a different metal that has a higher work function. On the graph above, where would the new line be by shining light at the replacement metal surface?

 

[etotheipi]

oh so the 2.5 is the work function i.e 2.5x10^-19J

Also @collinsmark what would happen if the metal is replaced with a different metal that has a higher work function. On the graph above, where would the new line be by shining light at the replacement metal surface?

How do you find the work function from the graph? Then, what would happen to the line if it were increased? What would happen to the gradient`?

Given that the equation of the line is $E_k = hf - \phi$?

 

[Jeff97]

How do you find the work function from the graph? Then, what would happen to the line if it were increased? What would happen to the gradient`?

Given that the equation of the line is $E_k = hf - \phi$?

I have just seen this now sorry. You find the work function from the Y-Intercept. The gradient would increase?

 

[etotheipi]

I have just seen this now sorry. You find the work function from the Y-Intercept. The gradient would increase?

Not quite. The gradient is still $h$. The result is that the line gets translated. Can you work out what this translation looks like?

 

[Jeff97]

So are you saying the line is just being moved higher with the same gradient?

 

[etotheipi]

So are you saying the line is just being moved higher with the same gradient?

You start off with the line $y = hf - \phi_{1}$. Now you've got the line $y = hf - \phi_{2}$ where $\phi_{2} > \phi_{1}$.

What would happen if I started with the line $y = 3x - 6$ and then also drew another line of $y = 3x - 10$ on the same graph? Try looking at what happens at $x=0$.

So if $\phi_{2} > \phi_{1}$, is the vertical translation positive or negative?

 

[Jeff97]

You start off with the line $y = hf - \phi_{1}$. Now you've got the line $y = hf - \phi_{2}$ where $\phi_{2} > \phi_{1}$.

What would happen if I started with the line $y = 3x - 6$ and then also drew another line of $y = 3x - 10$ on the same graph? Try looking at what happens at $x=0$.

So if $\phi_{2} > \phi_{1}$, is the vertical translation positive or negative?

The x intercept increases. Positive.

 

[Jeff97]

You start off with the line $y = hf - \phi_{1}$. Now you've got the line $y = hf - \phi_{2}$ where $\phi_{2} > \phi_{1}$.

What would happen if I started with the line $y = 3x - 6$ and then also drew another line of $y = 3x - 10$ on the same graph? Try looking at what happens at $x=0$.

So if $\phi_{2} > \phi_{1}$, is the vertical translation positive or negative?
[/QUOTE
am I right ?