stdevlin03
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A metal gives out photoelectrons that have a stopping voltage of2.6v
Will the light of a wavelength of 615 nm cause photoelectrons to be ejected?
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Hallouminatus
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TSR Physics forum does not provide "do my homework" service, so there are some guidelines that we expect everyone who is posting questions/problems to follow.
Before posting or asking your problems or questions, please attempt them first as you are required to show your attempt.

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stdevlin03
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(Original post by Hallouminatus)
TSR Physics forum does not provide "do my homework" service, so there are some guidelines that we expect everyone who is posting questions/problems to follow.
Before posting or asking your problems or questions, please attempt them first as you are required to show your attempt.

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1) A metal gives out photoelectrons that have a stopping voltage of2.6v

Will the light of a wavelength of 615 nm cause photoelectrons to be ejected?

2) I worked out the work function but am not sure what to do with it and it is negative which I'm not sure how to deal with
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Hallouminatus
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(Original post by stdevlin03)
1) A metal gives out photoelectrons that have a stopping voltage of2.6v

Will the light of a wavelength of 615 nm cause photoelectrons to be ejected?

2) I worked out the work function but am not sure what to do with it and it is negative which I'm not sure how to deal with
The graph on this page may help explain why work function is negative, and what to do with it. http://physicsnet.co.uk/a-level-phys...ectric-effect/
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stdevlin03
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(Original post by Hallouminatus)
The graph on this page may help explain why work function is negative, and what to do with it. http://physicsnet.co.uk/a-level-phys...ectric-effect/
so it can be negative it just means it is not enough to emit a photon from the surface?
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Hallouminatus
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Work function is always negative. Energy of photon + work function of metal = energy of emitted photoelectron. If the sum is negative, no electron can be emitted.
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stdevlin03
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(Original post by Hallouminatus)
Work function is always negative. Energy of photon + work function of metal = energy of emitted photoelectron. If the sum is negative, no electron can be emitted.
ok thanks
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stdevlin03
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(Original post by Hallouminatus)
Work function is always negative. Energy of photon + work function of metal = enesrgy of emitted photoelectron. If the sum is negative, no electron can be emitted.
so for this question how would that apply
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Hallouminatus
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(Original post by stdevlin03)
so for this question how would that apply
tbh, I'm a bit confused by this myself. Is there any other info or context?
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stdevlin03
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(Original post by Hallouminatus)
tbh, I'm a bit confused by this myself. Is there any other info or context?
A metal gives out photoelectrons that have stopping voltage of 2.6 v will light of wavelength cause photoelectrons to be ejected?

that is the exact question
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Sinnoh
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(Original post by stdevlin03)
A metal gives out photoelectrons that have stopping voltage of 2.6 v will light of wavelength cause photoelectrons to be ejected?

that is the exact question
From the stopping potential, you can work out the kinetic energy of the electrons.
From the wavelength, you can work out the energy of the incident light.
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stdevlin03
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(Original post by Sinnoh)
From the stopping potential, you can work out the kinetic energy of the electrons.
From the wavelength, you can work out the energy of the incident light.
so the Ek is 2.6x1.6*10^-19
and f = 3*10^8/ 615*10^-9

how do I answer the question from here?
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Sinnoh
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(Original post by stdevlin03)
so the Ek is 2.6x1.6*10^-19
and f = 3*10^8/ 615*10^-9

how do I answer the question from here?
Well you could just work out the photon energy directly as E = \dfrac{hc}{\lambda}. Is it higher or lower than the kinetic energy of the photoelectrons?

Does it mention the work function of the metal anywhere?
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Hallouminatus
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(Original post by Sinnoh)
From the stopping potential, you can work out the kinetic energy of the electrons.
From the wavelength, you can work out the energy of the incident light.
OK, I might be missing something really obvious here, or misreading the question, but it still looks like something is missing. To answer the question whether the 615 mm light will cause emission of a photoelectron, we need to know the work function, but the information that electrons are emitted at 2.6 eV is not sufficient to give the work function unless we know the frequency of the incident light.
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Sinnoh
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(Original post by Hallouminatus)
OK, I might be missing something really obvious here, or misreading the question, but it still looks like something is missing. To answer the question whether the 615 mm light will cause emission of a photoelectron, we need to know the work function, but the information that electrons are emitted at 2.6 eV is not sufficient to give the work function unless we know the frequency of the incident light.
Yeah it's a bit meaningless without the work function.

Although if the energy of the photons is less than the energy of the photoelectrons, then it can be answered because the answer is "definitely not".
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stdevlin03
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co

(Original post by Sinnoh)
Yeah it's a bit meaningless without the work function.

Although if the energy of the photons is less than the energy of the photoelectrons, then it can be answered because the answer is "definitely not".
Could you find the frequency using the wavelength and the threshold frequency and if it's lower it will not emit photoelectrons??
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Sinnoh
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(Original post by stdevlin03)
co


Could you find the frequency using the wavelength and the threshold frequency and if it's lower it will not emit photoelectrons??
All you need to do is compare the photon energy for a 615nm photon to the electron kinetic energy + work function
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stdevlin03
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(Original post by Sinnoh)
All you need to do is compare the photon energy for a 615nm photon to the electron kinetic energy + work function
where can I get the work function?
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stdevlin03
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(Original post by Sinnoh)
Well you could just work out the photon energy directly as E = \dfrac{hc}{\lambda}. Is it higher or lower than the kinetic energy of the photoelectrons?

Does it mention the work function of the metal anywhere?
can I find the work function using hf - phi = Ekmax
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Hallouminatus
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(Original post by stdevlin03)
can I find the work function using hf - phi = Ekmax
Yes, you could if you knew the frequency, but as I read the question, the light that caused the photoelectrons with a stopping voltage of 2,6 V is not the same as the 615 nm light in the second sentence: if it's the same, the question has already been answered and no calculation is needed because they've already said that electrons are emitted, but if it's different, then we don't know the wavelength or frequency so we can't calculate the work function.
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