(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Light of a wavelength 2000 Å falls on an aluminum surface. In aluminum, 4.2 eV are required to move an electron. What is the kinetic energy of

(a) the fastest, and

(b) the slowest emitted photoelectrons?

2. Relevant equations

[tex]K_{{\rm{max}}} = eV_0 [/tex]

3. The attempt at a solution

[tex]K_{{\rm{max}}} = eV_0 = - {\rm{1}}{\rm{.602}} \times 10^{ - 19} {\rm{C}} \times 4.2{\rm{eV}} \times {\rm{1}}{\rm{.602}} \times 10^{ - 19} \frac{{\rm{J}}}{{{\rm{eV}}}} = - 1.078 \times 10^{ - 37} {\rm{JC}}[/tex]

I doubt I'm doing this right, by using the required energy as the stopping potential. The units don't work out. But there's nothing in the book, except the formula I listed, that tells me how to do such a problem. The book gives me the impression that the stopping potential is experimentally derived. They give a graph for sodium, with points and a trendline, but it doesn't tell me what it is for aluminum (unless they buried it in an appendix). How do I do this problem?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Photoelectric effect and kinetic energy

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