Photoelectric Effect: Will Photoelectrons be Emitted?

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SUMMARY

The discussion centers on the photoelectric effect, specifically whether photoelectrons will be emitted when a metal with a work function of 9 x 10^-19 J is irradiated with UV radiation at 180 mm and 550 nm wavelengths. The calculations reveal that for 180 mm (0.18 m), the energy of the incident light is insufficient to release electrons, while for 550 nm, the calculations confirm no photoelectrons are emitted. The confusion arises from the incorrect wavelength unit for the first question, which should be in nanometers, indicating a potential typo in the homework statement.

PREREQUISITES
  • Understanding of the photoelectric effect and work function
  • Familiarity with the equations E = hf and E = hc/λ
  • Knowledge of electromagnetic spectrum and wavelength units
  • Basic principles of photon energy calculations
NEXT STEPS
  • Review the photoelectric effect and its dependence on wavelength
  • Learn about the electromagnetic spectrum and the significance of wavelength in photon energy
  • Practice calculations involving photon energy using E = hf and E = hc/λ
  • Investigate common misconceptions regarding wavelength units in physics problems
USEFUL FOR

Students studying physics, particularly those focusing on quantum mechanics and the photoelectric effect, as well as educators looking to clarify common errors in wavelength interpretation.

zebra1707
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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
 
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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?
 
The first question has the wavelength at 180 mm not nanometers. Convert that to meters is 0.18 m.

Cheers
 
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
 
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.
 
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
 

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