Cesium is used extensively in photocells and in television cameras because it ha

Question

Cesium is used extensively in photocells and in television cameras because it has the lower ionization energy of all the stable elements.

What is the maximum kinetic energy of a photoelectron ejected from cesium by 555nm light? Note that if the wavelength of the light used to irradicate the cesium surface becomes longer than 660nm, no photoelectrons are emitted.

I tried figuring this out by setting it as (6.6254x10^-34J/s)(3x10^8m/s)/(555nmx10^-9), and it's wrong according to the back of the book, since the answer is supposed to be 6.02x10^-20J.

Explanation / Answer

The answer would be the difference between the energy of a 555 nm photon and 660 nm photon, since the 660 nm photon establishes the work function of Cs.

E = hc/ so we want

hc/555 nm - hc/660 nm

= (6.626 * 10-34 J-s)(2.9979 * 108 m/s)(1/555 * 10-9 m - 1/660 * 10-9 m)

= 5.69 * 10-20 J which is not your book answer .

To get your book answer, the photon for which the photoelectrons are no longer emitted would have to have a wavelength of 667 or 668 nm. Are you sure you copied the 660 nm correctly?

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