Photoelectric Effect: Will Photoelectrons be Emitted?

In summary, the metal does not emit photoelectrons when irradiated with UV radiation of wavelength 180 mm or 550 nm.
  • #1
zebra1707
107
0

Homework Statement



Hi there. I have two questions and in both I believe that no photoelectrons will be emitted from the surface.

Question 1

A metal with a work function of 9 x 10^-19 J is irradiated with UV radiation of wavelength 180 mm. Will photoelectrons be emitted from the surface?

Homework Equations





The Attempt at a Solution




E = hf
f = c/λ

E = hc/λ

E = 6.63 x 10^-34 x 1.67 x 10^9
= 1.10 x 10^-24 J

The energy required to release the an electron from the metal is E0 = 9 x 10^-19

Therefore, the energy contained in the light, is insufficient to overcome the forces holding the electron to the surface of the metal.

The second question is that the same metal is used but this time irradiated by 550 nm

I come up with the following - which again indicates that nothing is released.

E = hf
f = c/λ

E = hc/λ

E = 6.63 x 10^-34 x 5.45 x 10^14
= 3.61 x 10^-19 J

Can someone please check, as the next question asks

Determine the maximum kinetic energy and velocity of any photoelectrons emitted in the above questions. But my calcs indicate that no photoelectrons are emitted. (The electrons rest mass is 9.1 x 10-31 kg.).


Cheers and thanks in advance
 
Physics news on Phys.org
  • #2
For 180nm light, electrons are emitted, you must have made an error in your calculation. I found the photon to have an energy of 1.1x10^(-18) J

For 550nm light, you seem to have done the calculation correctly. No electrons are emitted.

You know that smaller the wavelength, higher the energy of the photon right? How have you managed to calculate an energy for the 180nm light to be 5 orders of magnitude less than that of the 550nm light?
 
  • #3
The first question has the wavelength at 180 mm not nanometers. Convert that to meters is 0.18 m.

Cheers
 
  • #4
well ok! If that's the case then you're right on that one too... does make asking the electrons kinetic energy redundant. I've never known to be asked about the PE effect for anything other than nm wavelength light, it could be a typo in wherever the question came from and they did in fact mean nm? The whole question would make more sense if this was the case... :D
 
  • #5
JesseC said:
well ok! If that's the case then you're right on that one too... does make asking the electrons kinetic energy redundant. I've never known to be asked about the PE effect for anything other than nm wavelength light, it could be a typo in wherever the question came from and they did in fact mean nm? The whole question would make more sense if this was the case... :D

Electromagnetic radiation with a wavelength of 180mm would be described as microwave, not ultraviolet! Suspect a typo.
 
  • #6
I agree, it does making asking about the kinetic energy redundant - I think it is a typo..

Many thanks for all your help on this one.


JesseC said:
well ok! If that's the case then you're right on that one too... does make asking the electrons kinetic energy redundant. I've never known to be asked about the PE effect for anything other than nm wavelength light, it could be a typo in wherever the question came from and they did in fact mean nm? The whole question would make more sense if this was the case... :D
 

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which the emission of electrons from a material is induced by the absorption of electromagnetic radiation, such as light. It was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905 through his theory of the quantization of light.

2. How does the photoelectric effect work?

The photoelectric effect occurs when photons (particles of light) strike a material, causing its electrons to be ejected. Each photon must have a sufficient amount of energy, known as the threshold energy, to dislodge an electron from the material. The emitted electrons are called photoelectrons and their energy is equal to the difference between the photon's energy and the material's work function.

3. What factors affect the photoelectric effect?

The photoelectric effect is affected by the intensity and frequency of the incident light, as well as the type of material. Higher intensity light and shorter wavelengths (higher frequency) result in more photoelectrons being emitted. The work function of the material also plays a role, as it determines the amount of energy needed for electrons to be ejected.

4. What is the significance of the photoelectric effect?

The photoelectric effect has played a crucial role in the development of quantum mechanics and our understanding of the nature of light. It has also led to the invention of technologies such as solar cells, which use the photoelectric effect to convert light energy into electrical energy. Additionally, the photoelectric effect has been used to measure the energy of photons and to study the properties of different materials.

5. Can the photoelectric effect be explained by classical physics?

No, the photoelectric effect cannot be explained by classical physics. Classical physics predicted that the energy of the emitted electrons would increase with the intensity of the incident light, but this is not observed in experiments. It was not until Einstein proposed his theory of the quantization of light that the photoelectric effect could be fully explained.

Similar threads

  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
2
Replies
35
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Quantum Physics
Replies
12
Views
740
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top