Another photoelectric q! (1 Viewer)

[Premus]

this is from the success one book- q 8, But i dont really understand the answer.
i know that the intensity of light is proportional to the number of electrons...
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During an exp, a beam of uv light , of wavelength 200 nm , is incident on a target metal sheet. the UV light releases photoelectrons from the metal. the max kinetic energy of the photoelectrons was measured.

If a second beam of uv light, of equal intensity to the original uv beam, but of a shorter wavelength, was used in the exp, compare the photoelectrons released by the original beam with those produced when the second uv source is used.



Thanks!

 

[d_elmo]

well all i know is that the second beam has a shorter wavelength, so a lower frequency (im pretty sure...?) so this beam has lower energy photons compared to the first, and so the electrons liberated have less kinetic energy than those from the first beam... thats all i know. was that right?

 

[allstarr69]

doesnt a shorter wavelength mean higher frequency? c= f x wavelength

 

[d_elmo]

i dont remember, but i should... just reverse everything then. shorter wavelength=higher frequency=higher energy photons=electrons with more KE

 

[Premus]

Yeah thats what i thought....shorter wavelength => higher frequency

ok..the answer says

the frequency of the light has increased and this means that the energy of each photon is larger. the new beam will liberate photoelectrons with a greater maximum kinetic energy BUT because the intensity of the beam is equal to that of the original, the total energy per unit area is unchanged and, because each photon has a higher individual energy, there will be less photoelectrons being liberated from the surface than was the case with the original UV source.

????

 

[tempco]

and just mention that the intensity has no bearing on the KE of the photoelectrons.

BTW, does higher intensity = higher current?

 

[tempco]

the frequency of the light has increased and this means that the energy of each photon is larger.

E_photon = hf_photon

the new beam will liberate photoelectrons with a greater maximum kinetic energy

E_k = hf₀ - W, where W is the work function dependant on the target metal (W = hf₁ where f₁ is the threshold frequency of the metal). Therefore, the higher f₀, the more K_e.

BUT because the intensity of the beam is equal to that of the original, the total energy per unit area is unchanged and, because each photon has a higher individual energy, there will be less photoelectrons being liberated from the surface than was the case with the original UV source.

This is a bit iffy... one photon has the ability to free one photoelectron, as stated in quantum theory. And isn't the increase in frequency going to mean there is an increase in total energy per unit area?

BTW, Where did you get the answer from?

Sounds a bit like the black body radiation explanation though... werd.

 

[mr phil]

what the hell was that? haha im gonna fail phsycis

 

[allstarr69]

what the hell was that? haha im gonna fail phsycis

lol there is still time to study just nerd it from here on and use this site to your advantage and not your downfall

 

[ashtor]

yeah greater intensity means there is more charge flowing per unit time so there is higher current

 

[Premus]

ummmm i got it from success one. That was what the answer said :s

 

[tempco]

yeah greater intensity means there is more charge flowing per unit time so there is higher current

OK well, refering back to PrimusDog's post:

BUT because the intensity of the beam is equal to that of the original, the total energy per unit area is unchanged and, because each photon has a higher individual energy, there will be less photoelectrons being liberated from the surface than was the case with the original UV source.

When they say energy, or charge, is it the same thing? Refering to E = hf? or the charge of the electron?

 

[ashtor]

no thats mixed up...when your talking about energy = hf you're looking at the photons which is what Premus is talking about. The intensity of the light is constant, hence it ejects the same number of electrons that constitute the current. Thats why when you have more intense light you have greater current, since more electrons are being ejected (therefore therell be a greater flow of charge per unit time, which is the definition of current i think)

 

[tempco]

because each photon has a higher individual energy, there will be less photoelectrons being liberated from the surface than was the case with the original UV source.

So how does that make sense? The individual energy of photons is E = hf. The number of photoelectrons liberated is not related to that. How can the number of photoelectrons released reduce when photons have higher individual energy?

 

[snoopwogg]

ROFLMAO u ppl are just confusing yourselves and others. Go STUDY lmao. Higher intensity = more electrons liberated = higher current.. like ashtor said. and also shorter wavelenght = higher frequncy = more kinetic energy of the electrons = higher velocity of electron being liberated. Thats as complex as u gorra take it. i think ashtor is the only one here who knows wat he's talking about ... oh yeha and what nekkid os on about is right too..
That stuff that Premus got came out of a Excel book ay........ ^o)

 

[Premus]

sorry for the confusion...
so the answer in the book is wrong?

 

[Jase]

mm.. no the book is right. no wait.. im lost.

Okay so we know that, intensity doesn't change, so then the photocurrent cant change. But it does.. WHY??? I think it's because since the energy of each electron is higher, by some obscure 11th law of thermodynamics, the thermal distribution has to be balanced. There we go.

 

[mr EaZy]

i thought the equation for KE was:

KE= hf-W (w=work fn) not KE= W-hf?


any wayz do we need to know how to calculate the work function? coz i got no idea... isnt the stopping voltage measured so that we can know what the work function is?

 

[tempco]

That stuff that Premus got came out of a Excel book ay........ ^o)

That's the thing that worries me... Excel is reputable, and the questions they give may be similar to HSC questions the BOS can throw at us...

Okay so we know that, intensity doesn't change, so then the photocurrent cant change. But it does.. WHY???

Exactly... ?

 

[Jase]

The answer said, since the intensity stays the same, there is no effect on the current.
But then it says, since the energy is greater, there is less current.

So it's distribution of energy somehow.. yep...