Physics question 8 hw 21

Physics
Tutor: None Selected Time limit: 1 Day

Photons of what minimum frequency are required to remove electrons from gold? Note: The work function for gold is 4.8 eV.

Jul 15th, 2015

Thank you for the opportunity to help you with your question!

he photoelectric effect is when photons hit a material, usually a metal, and basically "knock" the electrons out of orbit by allowing it to absorb the energy the photon provides. So the minimum energy in one photon must be equal to the work function of the material, in this case gold.

First, we will turn the units of the work function from eV to J. 1eV is the amount of energy an electron gains when placed in a potential difference of 1V (hence the term electron-volt). This is found in the following way:

E=qV
=(1.60e-19)(1)
=1.60e-19 J

So 1 eV=1.60e-19J. We can use this to turn the work function into joules.

W=4.8eV * (1.60e-19 J/eV)
=7.68e-19 J

So the energy in each photon must be a MINIMUM of 7.68J. Thanks to some brilliant people at the beginning of last century, we know that the energy in each photon is its frequency multiplied by a value known as Planck's constant.

E0=hf
where E0 is the amount of energy in each photon, h is Planck's constant, and f is the frequency of the photon.

E0=hf
f=E0/h
=W/h
=7.68e-19 / 6.626e-34
=1.159e15Hz
=1.2e15Hz (after sig digs are taken into account)

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

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