# Modern physics - photon problems

Anyone know how to go about solving the following two problems?

- What is the minimum frequency photons need to knock an electron out of a copper target?

- If a beam of photons with 3 times the frequency of the photons from the first problem strikes a copper target, at what speed will the fastest electrons leave the copper?

Thank you.

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The Work Function of Copper

About 4.7eV. This means that a photon with a minimum frequency of 1.136 x 10^15 Hz {([4.7eV x 1.6022x10^(-19) J/eV] / h ) where h = 6.626 x 10^(-34) Js} is necessary. Energetic photon!

If a photon with a frequency = 3X the necessary frequency is incident, then the electron will be ejected with 9.4eV (3 x 4.7eV - 4.7eV) of K.E.

This corresponds to a speed of about 1.8 x 10^6 m/s. Approaching relativistic speeds...

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This problem has to do with the photoelectric effect.

The equation to use is

hv - W = K

h is planck's constant, v is the photon frequency. hv is the photon energy.

W is the work function of the metal, which is ~4.7 eV (7.5 x10^-19 J) for copper.

K is the kinetic energy.

In question 1 you are asking about a threshold energy, where are photon has just enough energy to kick the electron out of the copper, but give it no extra kinetic energy. K = 0 => hv = W

question 2, you are asking for the extra kinetic energy given a more energetic incident photon.

You can read more at http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html#c2

Cheyne

1)For your first question, you nedd to know abou the work function of copper, once you have that, you can equate it with hv where v=frequency of light striking the copper.

2)Now as per photoelectric effect , the total energy of the photon striking ..uses some of its energy to strike out the electron out of copper and rest conveerts into the KE of the ejected electron beam.

Therefore use hv=work Function+ Kinetic energy of electron Beam