Photoelectric Effect Graph and Work Function Questions

[AN630078]

Homework Statement

Hello, I have been reviewing the photoelectric effect when I came across the admittedly rather long problem below. I was keen to try to solve it since I had not used a graphical analysis for the photoelectric effect before, e.g. finding the work function in this way which has lead to some minor confusion. Could anyone offer any advice if I can improve upon my answers here, or generally dispense some light on the topic ?

Different stopping potentials were measured for different frequencies of electromagnetic radiation.

1. a. Plot the results from table 1 in a graph with maximum kinetic energy EK on the y-axis and frequency f on the x-axis.
b. Give the value of:
i. The threshold frequency
ii. The work function
c. Using Table 2 and your answer to b ii, find the metal that is being tested.
d. Using your graph and Table 3, approximately how much kinetic energy would an electron
emitted from an incident blue photon have?
e. What would happen if a beam of yellow light was shone on the metal?

Relevant Equations

1/2 m v^2 max = hf-Φ

1. a)I have plotted the graph on desmos and attached an image here.

b i. The threshold frequency is equal to the x-intercept ~ 5.6*10^14 Hz
ii. The work function is equal to the y-intercept ~ -3.75*10^19 J (would it be correct to state that this value is negative?)

c. Convert to eV;
3.75*10^19 J=2.34375 eV
Evaluating table 2 I believe that this approximately corresponds to the work function of sodium, 2.36 eV. Such differences in values may have arisen from graphical uncertainties on my part.

d. The photoelectric effect equation states;
1/2 m v^2 max = hf-Φ
Evaluating table 3 a blue photon has a frequency in range 606-668 THz. Converting to Hz this 6.06*10^14 Hz-6.68*10^14 Hz
Would the work function here be that I found with my graph ~ 3.75*10^19 J ?
1/2 m v^2 max =6.63*10^-34* 6.06*10^14-3.75*10^19
1/2 m v^2 max = -3.75 *10^19 J

1/2 m v^2 max =6.63*10^-34* 6.68*10^14-3.75*10^19
1/2 m v^2 max = -3.75 *10^19J

Which is the same for both extremes of the range of the blue photon's frequency. Would the value of the work function here be negative so the actual kinetic energy emitted from an incident blue photon would be positive i.e. 3.75 *10^19J?

e. If a beam of yellow light was shone on the metal the energy of the photon can be found by (frequency range of a yellow photon is 5.08*10^14 Hz-5.26*10^14
E=hf=6.63*10^-34*5.08*10^14=3.368*10^19
E=6.63*10^-34*5.26*10^14=3.487*10^19

Convert to eV; Energy in range= 2.10-2.12 eV

The work function of the sodium has been shown to be 2.36 in table 2. Since this energy is less than the work function, meaning the yellow photon does not have sufficient energy to eject an electron from the sodium metal.

 

[kuruman]

(a) Not OK, what did you plot? There are no labels to the axes.
(b) If you use the equation 1/2 m v^2 max = hf-Φ, it would not be correct to substitute a negative number for Φ. Do you see why?
(c) OK.
(d) This is clearly wrong. A kinetic energy is never negative. See my question in (b). I would forget about Joules and convert all energies to eV. It would be much easier to see what's going on.

 

[AN630078]

(a) Not OK, what did you plot? There are no labels to the axes.
(b) If you use the equation 1/2 m v^2 max = hf-Φ, it would not be correct to substitute a negative number for Φ. Do you see why?
(c) OK.
(d) This is clearly wrong. A kinetic energy is never negative. See my question in (b). I would forget about Joules and convert all energies to eV. It would be much easier to see what's going on.

Thank you for your reply.
a) I just plotted this quickly on desmos. I will attach my hand-drawn graph which features fully labelled axis.
b) Would it be wrong to substitute a negative number since the formula is subtracting the work function, so two negatives would make a positive, meaning it was no longer being subtracted but added?
d) The work function is negative on the graph, is this because negative energy is energy of absorption? The photoelectric surface must absorb a certain amount of energy in order to emit photoelectrons: ie the Work function.
The value of the work function in eV = 2.36 eV
Would that mean;
1/2 m v^2 max =6.63*10^-34* 6.06*10^14-(-2.36 eV)
1/2 m v^2 max = 2.36 J

1/2 m v^2 max =6.63*10^-34* 6.68*10^14-(-2.36 eV)
1/2 m v^2 max = 2.36 J

Is this still incorrect?

Would my answer for e be wrong ?

 

[kuruman]

Usually, one writes the photoelectric equation as $$h\nu=\frac{1}{2}mv^2+\phi$$ saying that the energy of the photon is split into what it costs the electron to break free (work function) plus the leftover kinetic energy of the electron. In this expression the work function is a positive number. If you want to find the kinetic energy, you have to move the work function to the other side of the equation and change its sign. The work function is a positive number. The fact that the intercept in your (still unlabeled) plot is negative does not mean that the work function itself is negative. Assuming that the horizontal axis is frequency, and the vertical axis the stopping voltage, how do you think the negative intercept is to be interpreted?

 

[AN630078]

Usually, one writes the photoelectric equation as $$h\nu=\frac{1}{2}mv^2+\phi$$ saying that the energy of the photon is split into what it costs the electron to break free (work function) plus the leftover kinetic energy of the electron. In this expression the work function is a positive number. If you want to find the kinetic energy, you have to move the work function to the other side of the equation and change its sign. The work function is a positive number. The fact that the intercept in your (still unlabeled) plot is negative does not mean that the work function itself is negative. Assuming that the horizontal axis is frequency, and the vertical axis the stopping voltage, how do you think the negative intercept is to be interpreted?

Thank you for your reply. I was not familiar with this format of the photoelectric equation. The vertical axis is maximum kinetic energy and the horizontal axis is the frequency in my graph. Does this effect the work function than if the vertical axis were the stopping voltage?

 

[kuruman]

Thank you for your reply. I was not familiar with this format of the photoelectric equation. The vertical axis is maximum kinetic energy and the horizontal axis is the frequency in my graph. Does this effect the work function than if the vertical axis were the stopping voltage?

No, the vertical axis is not kinetic energy unless you measured the speed of each electron, a daunting task to say the least. What did you measure and how is that related to the kinetic energy of the electrons?

 

[AN630078]

No, the vertical axis is not kinetic energy unless you measured the speed of each electron, a daunting task to say the least. What did you measure and how is that related to the kinetic energy of the electrons?

No question 1. a. states " Plot the results from table 1 in a graph with maximum kinetic energy EK on the y-axis and frequency f on the x-axis"

 

[kuruman]

And exactly how did you measure this maximum kinetic energy? Did you read an instrument that had a display labeled "kinetic energy"?

 

[AN630078]

And exactly how did you measure this maximum kinetic energy? Did you read an instrument that had a display labeled "kinetic energy"?

No, I did not measure the kinetic energy, I calculated it using 1/2 mv^2=e*Vs. I understand this question is referring to the photoelectric cell experiment but it does specify to plot maximum kinetic energy on the y-axis, not the stopping voltage.

 

[AN630078]

Sorry I an confused should I not have plotted maximum kinetic energy on the y-axis?

 

[AN630078]

And exactly how did you measure this maximum kinetic energy? Did you read an instrument that had a display labeled "kinetic energy"?

d) Do you mean convert Plancks's constant into eV also;
6.63*10^19 J = 4.1357*10^-15 eV

1/2 m v^2 max =4.1357*10^-15* 6.06*10^14-2.36 eV
1/2 m v^2 max =0.1462342 ~ 1.46 * 10 ^-1 J

1/2 m v^2 max =4.1357*10^-15* 6.68*10^14-2.36 eV
1/2 m v^2 max = 0.4026476 ~ 4.03*10^-1 J

So the maximum kinetic energy a phototelectron emitted from an incident blue photon would have would be between 1.46 * 10 ^-1 J-4.03*10^-1 J
Or in eV; 9.13*10^17eV -2.51*10^18 eV

Would this be incorrect?

 

[AN630078]

Sorry I am still confused?

 

[kuruman]

Sorry I an confused should I not have plotted maximum kinetic energy on the y-axis?

Since you were asked to plot kinetic energy on the y-axis, that's what you should do. I am not sure you understand how to get the work function from the plot. The straight line equation has the general form $y=ax+b$. Can you identify for me what these are in terms of the photoelectric wave equation?

In part (d) perhaps I did not explain myself clearly. Forget Joules. Find the energy of the blue photon in eV. If you subtract from it the work function, which you already have in eV, what do you get?

 

[AN630078]

Since you were asked to plot kinetic energy on the y-axis, that's what you should do. I am not sure you understand how to get the work function from the plot. The straight line equation has the general form $y=ax+b$. Can you identify for me what these are in terms of the photoelectric wave equation?

In part (d) perhaps I did not explain myself clearly. Forget Joules. Find the energy of the blue photon in eV. If you subtract from it the work function, which you already have in eV, what do you get?

Thank you for your reply. I learned about the photoelectric cell experiment from my textbook which states the y-intercept will represent the value of the work function.
Comparing the photoelectric effect equation with the equation of a straight line;
1/2mv^2=hf-Φ
y=mx+c
Then when the value of y=0 (the x-intercept) this value represents the threshold frequency of for the anode metal.

For d) Finding the energy of the incident blue photon in the range of the given frequency;

1/2 m v^2 max =4.1357*10^-15 eV* 6.06*10^14-2.36 eV
1/2 m v^2 max =0.1462342 ~ 1.46 * 10 ^-1 eV

1/2 m v^2 max =4.1357*10^-15eV* 6.68*10^14-2.36 eV
1/2 m v^2 max = 0.4026476 ~ 4.03*10^-1 eV

Is this correct (Planck's constant and the work function are in eV and the frequency in Hz) so the maximum kinetic energy a phototelectron emitted from an incident blue photon would have would be between 1.46 * 10 ^-1 eV-4.03*10^-1 eV

 

[kuruman]

Your range of kinetic energies for blue light looks about right.

 

[AN630078]

Your range of kinetic energies for blue light looks about right.

Really? Thank you for your help . Are my other solutions correct now also or do I need to amend further details?

 

[kuruman]

You absolutely need to fix your plot. The labels and numbers are not the only issue with it. If you label the vertical axis as "kinetic energy", you need to explain what the negative "kinetic energy" at the y-intercept means and why.

 

[AN630078]

You absolutely need to fix your plot. The labels and numbers are not the only issue with it. If you label the vertical axis as "kinetic energy", you need to explain what the negative "kinetic energy" at the y-intercept means and why.

Thank you for your reply. I will try to amend my plot. Moreover, if the question further asked;
For an incident radiation with a wavelength of 300 nm:
1. What is the velocity of the emitted electron?
2. i. What is the de Broglie wavelength of the electron?
ii. What effect would increasing the intensity of the beam have?

To find the velocity of the emitted photoelectron would I rearrange the photoelectric equation in terms of v;

v=√2( hf-Φ)/m
Therefore, working in eV:
v=√2((4.1357*10^-15 eV * 3*10^8 ms^1/3.0*10^-7 m)-2.36)/9.1*10^-31 kg
v=√2(4.1357-2.36)/9.1*10^-31
v=1.975509... *10^15 ~ 1.98 *10^15 ms^-1

2i. the de Broglie equation states; electron wavelength=Planck's constant/momentum
λ=h/p
λ=6.63*10^-34 J/(1.98 *10^15 ms^-1 * 9.1*10^-31 kg)
λ=6.63*10^-34 J/1.78 *1-^-15 kg ms^-1
λ=3.6880 ... * 10^-19 ~ 3.69 *10^-19 m

ii. By increasing the intensity of the beam would this increase the number of photons in the beam and thus increases the number of electrons excited? Moreover, would the frequency of the incident beam and therefore the maximum kinetic energy of the emitted photoelectrons also increase?

 

[kuruman]

The method for parts 1 and 2(i) are correct - I did not check your numbers.

ii. By increasing the intensity of the beam would this increase the number of photons in the beam and thus increases the number of electrons excited?

Yes.

Moreover, would the frequency of the incident beam and therefore the maximum kinetic energy of the emitted photoelectrons also increase?

What equation is there that relates the frequency of 300 nm photons to the beam intensity, i.e. how many you have? What about the maximum KE, what equation is there that relates it to intensity?

 

[AN630078]

The method for parts 1 and 2(i) are correct - I did not check your numbers.


Yes.

What equation is there that relates the frequency of 300 nm photons to the beam intensity, i.e. how many you have? What about the maximum KE, what equation is there that relates it to intensity?

Thank you for your reply. I have checked my calculations a few times for parts 1 and 2i, and think they are correct. Do you think the velocity would be too high?

2.ii I know that the equation for intensity is I= power/area, but I do not really think that is applicable here. I actually have been thinking and believe that the kinetic energy is independent of the intensity of the incident beam of radiation. However, I think that the number of photoelectrons ejected will increase at higher intensities, since high-intensity EM radiation consists of a greater numbersof photons per unit area. Furthermore, all of these photons having the same characteristic energy = hf.

 

[kuruman]

Thank you for your reply. I have checked my calculations a few times for parts 1 and 2i, and think they are correct. Do you think the velocity would be too high?

Yes, it is too high. It is of order 10¹⁵ m/s. By comparison, the speed of light is 3x10⁸ m/s. That should have tipped you off.

2.ii I know that the equation for intensity is I= power/area, but I do not really think that is applicable here. I actually have been thinking and believe that the kinetic energy is independent of the intensity of the incident beam of radiation. However, I think that the number of photoelectrons ejected will increase at higher intensities, since high-intensity EM radiation consists of a greater numbersof photons per unit area. Furthermore, all of these photons having the same characteristic energy = hf.

Yes.

 

[AN630078]

Yes, it is too high. It is of order 10¹⁵ m/s. By comparison, the speed of light is 3x10⁸ m/s. That should have tipped you off.

Yes.

Thank you for your reply, yes that is what I thought. Where have I gone wrong in my calculations?
I rearranged the photoelectric effect equation in terms of velocity:
v=√2( hf-Φ)/m
And continued to substitute in the known values, converting the work function and h into eV. Would it be correct to use the same work function here for sodium, since I assume it is the same metal being tested?
Should I have left Planck's constant in joules and converted the work function to joules?

v=√2((6.63*10^-34 J * 3*10^8 ms^-1/3.0*10^-7 m)-3.78*10^-19 J)/9.1*10^-31 kg
v=√2(6.63*10^-19-3.78*10^-19)/9.1*10^-31
v=791003.198 ~ 7.9 *10^5 ms^-1 ?

 

[kuruman]

Yes, you need to have all quantities in SI units for this part. This means inserting (hf - Φ) to the equation in Joules. If you know how to use a spreadsheet, I would strongly recommend that you do so. Not only it makes nice plots for you, but also you need to enter your input parameters and formulas only once. If you make a mistake it would be easy to find and fix. Furthermore, knowing how to use a spreadsheet is, in my opinion, as useful a skill as knowing how to use a word processor especially if you are in a technical field.

 

[AN630078]

Yes, you need to have all quantities in SI units for this part. This means inserting (hf - Φ) to the equation in Joules. If you know how to use a spreadsheet, I would strongly recommend that you do so. Not only it makes nice plots for you, but also you need to enter your input parameters and formulas only once. If you make a mistake it would be easy to find and fix. Furthermore, knowing how to use a spreadsheet is, in my opinion, as useful a skill as knowing how to use a word processor especially if you are in a technical field.

Thank you very much for your reply I tremendously appreciate all of your help it has been invaluable. I am mildly familiar with using spreadsheets but it is admittedly a dexterity which I am yet to develop fully.
Moreover, I have attached my hand-drawn plot, would this be better? I know that the work function is negative at the y-intercept but would this be ok since comparing the photoelectric effect equation with the equation of a straight line;
Ek=hf-Φ
y=mx+c

Therefore, by subtracting the work function in the photoelectric effect equation, its negative value will cancel to a positive?

 

[kuruman]

I know that the work function is negative at the y-intercept ...

The work function is never negative, so don't say that. I know what you mean to say and this is how you say it. When one compares the equations Ek=hf-Φ and y=mx+c, one sees that the y-intercept c in a plot of Ek vs. f is the negative of the work function Φ. In this case c = - 2.36 eV, therefore Φ = +2.36 eV. And yes, this new plot is better.

 

[AN630078]

The work function is never negative, so don't say that. I know what you mean to say and this is how you say it. When one compares the equations Ek=hf-Φ and y=mx+c, one sees that the y-intercept c in a plot of Ek vs. f is the negative of the work function Φ. In this case c = - 2.36 eV, therefore Φ = +2.36 eV. And yes, this new plot is better.

Thank you very much for your reply and for offering a preferable statement