Photoelectric Example: Difference between revisions

Latest revision as of 13:19, 18 January 2014

The work function is

$eW_{0}=h\nu _{th}={\frac {hc}{\lambda }}={\frac {12400{\text{ eV}}\cdot \mathrm {\AA} }{6000\,\mathrm {\AA} }}=2.07{\text{ eV}}$

The photoelectric equation then gives

$eV_{s}=h\nu -eW_{0}={\frac {hc}{\lambda }}-eW_{0}$

$2.5{\text{ eV}}={\frac {12400{\text{ eV}}\cdot \mathrm {\AA} }{\lambda }}-2.07{\text{ eV}}\Rightarrow \lambda =2713\,\mathrm {\AA}$

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-''Source:'' "Theory and problems of Modern Physics", Ronald Gautreau,Problem 9.13
+The work function is
-'''Problem:''' The emitter in a photoelectric tube has a threshold wavelength of 6000 Å. Determine the wavelength of the light incident on the tube if the stopping potential for this light is 2.5 V.
-'''Solution:''' The work function is
 <math>eW_{0}=h\nu_{th}=\frac{hc}{\lambda }=\frac{12400 \text{ eV}\cdot\AA}{6000\,\AA}=2.07\text{ eV}</math>
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 <math>2.5\text{ eV}=\frac{12400 \text{ eV}\cdot\AA}{\lambda }-2.07\text{ eV}\Rightarrow \lambda =2713\,\AA</math>
+Back to [[Photoelectric Effect#Problem|Photoelectric Effect]]