Physics question 8 hw 21
Physics

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Photons of what minimum frequency are required to remove electrons from gold? Note: The work function for gold is 4.8 eV.
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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 electronvolt). This is found in the
following way:
E=qV
=(1.60e19)(1)
=1.60e19 J
So 1 eV=1.60e19J. We can use this to turn the work function into joules.
W=4.8eV * (1.60e19 J/eV)
=7.68e19 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
=7.68e19 / 6.626e34
=1.159e15Hz
=1.2e15Hz (after sig digs are taken into account)
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