Colour of light producing photoelectrons?

[Lauren12]

Homework Statement

a) calculate the energy, in joules, of a quantum of light with a wavelength of 590 nm.
b)will this colour be able to produce photoelectrons from the surface of a metal with a work function of 3.0 eV? Explain your reasoning.

Homework Equations

f=c/λ
E=fh
ek=hf-W

The Attempt at a Solution

a)
590 nm = 5.9*10^-7 m

f=c/λ
f=c/(5.9*10^-7)
f= 5.085*10^14

E=fh
E=5.085*10^14*h
E= 3.3712*10^-19 J

b)
Ek will be zero for minimum kinetic energy.
W=3.0eV = 4.80653199*10^-19 J

ek=hf-W
W=hf
f=W/h
f=4.80653199*10^-19 J/h
f= 7.25 * 10^14
Therefore this colour of light will not be able to produce photoelectrons from the surface of metal because the minimum frequency to produce light is larger than the frequency of the light.

If someone would confirm that what I have done is correct or not that would be great! I'm not sure if I'm even on the right track! Thanks in advance!

 

[ehild]

It is correct, but you could have concluded the answer from the energy of the photon, which is less than the work function, that it could not produce photoelectrons.

ehild

 

[TSny]

Your process looks good. Note that you can answer part (b) by just comparing your answer for the energy of the photon from (a) with the work function energy. That way, you don't need to find the frequency corresponding to the work function. But your way is also good.

[EDIT: Looks like echild beat me to the punch (this time).]

 

[Lauren12]

Thank you both! very much appreciated :D

 

[sophiecentaur]

It's 'Wavelength or Frequency' that governs the production of photo-electrons and not "colour". Colour is merely the brain's interpretation to various combinations of different wavelenths. This is highly relevant if you imagine a surface with a work function just higher than that of the Sodium D ('yellow') line. There will be no photo-electrons produced when light from a sodium lamp arrives. However, a matching 'Yellow' colour can easily be produced by a combination of long wavelength light ('spectral red') and mid wavelength light ('spectral Green'). The spectral green component of the yellow light will have sufficient energy to produce electrons.
So it's best to avoid the word 'colour' unless you are talking about human eyes.

 

[ehild]

You are right, colour is not appropriate in case of photo electricity, but it was all right in the context of this problem. It was about a quantum of light of a specific colour, and a quantum of light has a definite frequency. A mixed colour involves different frequencies, so more than one kind of photons.

ehild

 

[sophiecentaur]

You are right, colour is not appropriate in case of photo electricity, but it was all right in the context of this problem. It was about a quantum of light of a specific colour, and a quantum of light has a definite frequency. A mixed colour involves different frequencies, so more than one kind of photons.

ehild

Fair comment about an individual photon having only one frequency. I still hold that using the term 'colour' is really inappropriate. It is a small step from this, which is only a form of mis-use to being actually wrong so it is best to avoid using the term at all. Your defense goes with the 'letter' of the law and not the spirit, I think. You are not a member of the bar, by any chance?

 

[ehild]

Colour of photon was not my idea, but it appeared in the text of the problem. And such questions do appear in schools if you can get photoelectric effect with red light, and the students have to answer.

By the way, what bar do you mean? ehild

 

[sophiecentaur]

Colour of photon was not my idea, but it appeared in the text of the problem. And such questions do appear in schools if you can get photoelectric effect with red light, and the students have to answer.

By the way, what bar do you mean?


ehild

Questions that are set in public exams are not always very well worded and some of them are serious nonsense. I doubt that many examiners are too familiar with colourimetry and they try to make questions appear friendly and approachable. I could give you some very red-looking light that would contain a significant amount of 'blue-end' photons. That's my point.

The Bar - a British Court of Law where you would find a barrister. They are paid to arguefy better than the other guy and 'get a result'. A very starchy affair compared with the US style - if TV films are anything to go by.