Photons striking a metal target

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Homework Statement



Sodium, aluminium and silver have work functions of 2.46, 4.08 and 4.73 eV respectively. Which, if any of these metals will emit photoelectrons when irradiated by Balmer-B photons from a hydrogen discharge lamp?


Homework Equations



[tex]Energy = 13.6(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}[/tex]

The Attempt at a Solution



Using the above formula I have calculated that Balmer-B photons from a hydrogen discharge lamp have an energy of 2.55 eV.

My answer is none though, because I was under the impression that a photon has to give up all or none of its energy in a collision like this. So wouldn't the work function have to exactly match the photon energy?

That doesn't really sound right, but I'm sure that there is a condition about photons giving up all or none of their energy in this photoelectric stuff. Can someone please clarify?
 
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Andrusko said:

Homework Statement



Sodium, aluminium and silver have work functions of 2.46, 4.08 and 4.73 eV respectively. Which, if any of these metals will emit photoelectrons when irradiated by Balmer-B photons from a hydrogen discharge lamp?


Homework Equations



[tex]Energy = 13.6(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}[/tex]

The Attempt at a Solution



Using the above formula I have calculated that Balmer-B photons from a hydrogen discharge lamp have an energy of 2.55 eV.

My answer is none though, because I was under the impression that a photon has to give up all or none of its energy in a collision like this. So wouldn't the work function have to exactly match the photon energy?

That doesn't really sound right, but I'm sure that there is a condition about photons giving up all or none of their energy in this photoelectric stuff. Can someone please clarify?
The photon must indeed 'give up all or none' of it's energy to the electrons, but this doesn't mean that the photon energy needs to be equal to the work function. The work function is the energy required for the electron to become 'unbound' from the surface of the metal, any additional energy imparted on the electron by a photon contributes to the kinetic energy of the electron.

Does that make sense?
 
Ah, of course! I see now. Thankyou very much!