If an electron in metallic cesium absorbs a photon of red light (specifically, =

Question

If an electron in metallic cesium absorbs a photon of red light (specifically, =6.6x10-7 meters in a vacuum), all the energy is used up in escaping from the attractive forces in the metal. There is no energy left over after the electron has gotten out (just barely). Suppose blue-green light (with =5x10-7 meters) is used. What percentage of the photon's energy does the electron retain after it has escaped from the metal?

(Recall that a photon's energy is determined by the frequency of the light. You can, however, readily convert wavelength information into frequency information. If you do some algebra, you will find that Planck's constant cancels out in the final answer. That will save you some nasty multiplying - and you will see things more clearly.)

Explanation / Answer

work function = W = 1240/660 = 1.88 eV

Energy of photon = hf = 1240/500 = 2.48 eV

From Photoelectic effect, KE of ejected electrons = hf - W = 0.6 eV

The percentage of the photon's energy does the electron retain after it has escaped from the metal

= (0.6/2.48)*100 = 24.2%

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