## photoelectric effect

$c=\lambda v$

Tiffany Cao 1D
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### photoelectric effect

In the function, Ek=hv-(work function), how would we solve for the work function? In addition, if the work function exceeds the kinetic energy, does this mean the photon is not able to be emitted from the piece of metal?

Chem_Mod
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### Re: photoelectric effect

Hi Tiffany,

The work function is a constant that depends on the piece of metal from which electrons are ejected and represents the energy threshold at which electrons can be ejected from this metal. Elight$=h*\nu$ depends on the type of light shown on the metal piece (for example, UV light has more energy per photon than IR light does). Whether or not there are electrons ejected and if they have kinetic energy depends on the energy of the light shown on the metal piece and the work function. If Elight$=h*\nu$ is less than the work function, then there are no electrons ejected (and thus no kinetic energy since negative kinetic energy is impossible). When Elight=$h*\nu$ equals work function, electrons are barely ejected and have no kinetic energy. When Elight=$h*\nu$ is greater than work function, electrons are ejected and have kinetic energy.

Hope this helps!

Sarah_Stay_1D
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### Re: photoelectric effect

For this class the work function would be given to you. You would only have to solve for the work function if you had all other variables except the work function.

Kinetic energy is different from the energy of a photon. The kinetic energy is the excess energy released from the ejected electron. hv represents the energy supplied by a photon, so the product of h and v must be greater than the work function in order for the electron to be ejected.