Photoelectric Example: Difference between revisions

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The photoelectric equation then gives
The photoelectric equation then gives


<math>ev_{s}=h\nu-eW_{0}=\frac{hc}{\lambda}-eW_{0}</math>
<math>eV_{s}=h\nu-eW_{0}=\frac{hc}{\lambda}-eW_{0}</math>


<math>2.5\text{ eV}=\frac{1.24\times 10^3 \text{ eV}\cdot\AA}{\lambda }-2.07\text{ eV}\Rightarrow \lambda =2713\,\AA</math>
<math>2.5\text{ eV}=\frac{1.24\times 10^3 \text{ eV}\cdot\AA}{\lambda }-2.07\text{ eV}\Rightarrow \lambda =2713\,\AA</math>

Revision as of 14:40, 18 March 2013

Source: "Theory and problems of Modern Physics", Ronald Gautreau,Problem 9.13

Problem: The emitter in a photoelectric tube has a threshold wavelength of . 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

The photoelectric equation then gives