A certain minimum energy(E min) is required to eject an electron from a photosen

Question

A certain minimum energy(E min) is required to eject an electron from a photosensitive surface. Any energy obsorbed beyond this minimum gives kinetic energy to the ejected electron. When 540-nm light falls on a cesium surface, an electron is ejected with a kinetic energy of 6.69 X 10 ^-20 J. When the wavelength is 400 nm, the kinetic energy is 1.96 X 10 ^-19 J.

A) Calculate the minimum energy (E min) for cesium, in joules.

B) Calculate the longest wavelength, in nanometers, that will eject an electron from cesium.

Explanation / Answer

The energy of a photon is:

E = hc / ?

The energy of a photon with 540nm wavelength is therefore:

E = 6.626068*10^-34 * 299792458 / (540*10^-9)

E = 3.6786*10^-19 J

Since the kinetic energy of the electron is known:

Emin = 3.6786*10^-19 - 2.60*10^-20

Emin = 3.4186*10^-19

Emin = 3.42*10^-19 J

The longest wavelength of light that will cause an ejection of an electron has an energy equal to that of Emin:

Emin = hc / ?

3.42*10^-19 = 6.626068*10^-34 * 299792458 / ?
? = 5.81*10^-7 m

? = 581 nm

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