# Chemistry

Light with a wavelength of 425 nm fell on a potassium surface, and electrons were ejected at a speed of 4.88 X 10^5 m/s. What energy was expended in removing an electron from the metal? Express the answer in joules (per electron) and in kilojoules per mole ( of electrons).

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1. energy of the photon = hc/lambda.
You know h,c, and lambda. Be sure to change lambda to meters.
KE of the ejected electron
= 1/2 mv2. Remember to use kg for mass of electron.
Energy expended is the difference of the two. This will be per electron ejected and it will be in joules. Multiply that by 6.02 x 10^23 to change to per mol of electrons.

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posted by DrBob222

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