How to approach this photoelectric effect question

• Safder Aree
In summary: You're right, it doesn't make sense it takes less time to collect more energy with less surface area. Would swapping the ratio be the correct method? So taking Total Area/area of the atom?
Safder Aree
I am currently taking my first Quantum Mechanics course and was given this problem in a practice set(we are supposed to refer to old intro textbooks). We haven't covered the photoelectric effect (just theory) much in class and reading through other textbooks, I wasn't able to find any similar questions. Any hints on how to approach this question would be greatly appreciated. Thank you.

1. Homework Statement
https://i.imgur.com/1FT6skF.jpg

KE = hf - phi

The Attempt at a Solution

I know phi, so KE = hf - 2.3eV
However, I do not know how to find f since I don't know wave length.
I also do not understand how I can relate it back to time.

Attachments

37.4 KB · Views: 300
Last edited by a moderator:
This problem is asking you to perform a classical calculation. That means that you should do it without recourse to the photoelectric formula which, as you already found out, is not useful here. You know the incident power, so find how long it will take for 2.3 eV to be collected.

kuruman said:
This problem is asking you to perform a classical calculation. That means that you should do it without recourse to the photoelectric formula which, as you already found out, is not useful here. You know the incident power, so find how long it will take for 2.3 eV to be collected.

Would this just be P = W/T? Then solving for T in this instance?

It would be that. Be careful how you use the numbers and watch the units that are given to you.

kuruman said:
It would be that. Be careful how you use the numbers and watch the units that are given to you.
So
2.3ev in Joules = 3.68501*10^-19

1.6*10^-9 = W/T
T = 1.6*10^9/3.68501*10^-19
T=2.303*10^-10

The units seems to work out if I did dimensional analysis right. If this right?

Also, please put units next to numbers. What kind of units should W have in the equation P = W/T? What kind of units does 1.6×10-9 have as given in the question?

Safder Aree
kuruman said:
Also, please put units next to numbers. What kind of units should W have in the equation P = W/T? What kind of units does 1.6×10-9 have as given in the question?

Okay I'll write in the units.

So Power is in watts (J/S)

P = 1.6*10^-9 J/s
W= 2.3eV = 3.68501*10^-19 Joules

Therefore,

P= W/T
T = W/P

T = (1.6*10^9 J/s) /(3.68501*10^-19 J)
T=2.303*10^-10s

That's the time required for the entire area to collect 2.3 eV because P is the area of the cathode. Is that the time that the question is asking you to find?

Safder Aree
kuruman said:
That's the time required for the entire area to collect 2.3 eV because P is the area of the cathode. Is that the time that the question is asking you to find?

No it's asking for it over the area of one atom. So this is my strategy to find that:

Find the area of the atom (assuming 2d dimensions):
A= pi*r^2
A = 3.14159*10^-18m^2

Then multiply the time by the ratio of A/Total Area:

=2.303*10^-10s * (3.14159*10^-18 m^2)/(1*10^-3 m^2)
=7.235*10^-25

Safder Aree said:
No it's asking for it over the area of one atom. So this is my strategy to find that:

Find the area of the atom (assuming 2d dimensions):
A= pi*r^2
A = 3.14159*10^-18m^2

Then multiply the time by the ratio of A/Total Area:

=2.303*10^-10s * (3.14159*10^-18 m^2)/(1*10^-3 m^2)
=7.235*10^-25
I assume your numbers are given in seconds. Please avoid calculating numbers without trying to figure out if your answers make sense. Here is a question I have: If it takes T=2.303*10^-10s for the entire area to collect 2.3 eV, does it make sense that an atom, which has a much much smaller area, will collect the same amount of energy in only 7.235*10^-25 s?

kuruman said:
I assume your numbers are given in seconds. Please avoid calculating numbers without trying to figure out if your answers make sense. Here is a question I have: If it takes T=2.303*10^-10s for the entire area to collect 2.3 eV, does it make sense that an atom, which has a much much smaller area, will collect the same amount of energy in only 7.235*10^-25 s?

You're right, it doesn't make sense it takes less time to collect more energy with less surface area. Would swapping the ratio be the correct method? So taking Total Area/area of the atom?

What do you think? Don't guess, but reason it out. If the total area of 1 mm2 receives 1.6×10-9 W, what fraction of that power does one atom receive given the radius of one atom?

kuruman said:
What do you think? Don't guess, but reason it out. If the total area of 1 mm2 receives 1.6×10-9 W, what fraction of that power does one atom receive given the radius of one atom?
This makes sense now to me. I was confused about the how to approach it. Thank you for all the help!

What is the photoelectric effect?

The photoelectric effect is a phenomenon where electrons are emitted from a material when it is exposed to light. This was first observed by Heinrich Hertz in 1887 and further studied by Albert Einstein in 1905.

How do you approach a photoelectric effect question?

The first step is to carefully read and understand the question. Then, make sure you have a solid understanding of the basic concepts and equations related to the photoelectric effect. Next, identify what information is given and what is being asked in the question. Finally, use the relevant equations and principles to solve the problem.

What are the key equations related to the photoelectric effect?

The key equations are the photoelectric effect equation (E = hf - Φ), the work function equation (Φ = hf0), and the energy of a photon equation (E = hf).

What are the main factors that affect the photoelectric effect?

The main factors are the intensity of light, the frequency of light, and the work function of the material. Increasing the intensity or frequency of light will increase the number of electrons emitted, while a higher work function will require more energy to eject electrons.

How does the photoelectric effect support the particle theory of light?

The photoelectric effect provides evidence for the particle theory of light by showing that light behaves as discrete packets of energy (photons) rather than as a continuous wave. This is demonstrated by the fact that increasing the frequency of light increases the kinetic energy of the emitted electrons, which is consistent with the equation E = hf.

• Introductory Physics Homework Help
Replies
35
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Quantum Physics
Replies
5
Views
716
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Quantum Physics
Replies
13
Views
1K