Photoelectric Example: Difference between revisions

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'''Solution:''' The work function is  
'''Solution:''' The work function is  


<math>eW_{0}=h\nu_{th}=\frac{hc}{\lambda }=\frac{1240\times 10^3 \text{ eV}\cdot\AA}{6000\,\AA}=2.07\text{ eV}</math>
<math>eW_{0}=h\nu_{th}=\frac{hc}{\lambda }=\frac{12400 \text{ eV}\cdot\AA}{6000\,\AA}=2.07\text{ eV}</math>


The photoelectric equation then gives
The photoelectric equation then gives
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<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{12400 \text{ eV}\cdot\AA}{\lambda }-2.07\text{ eV}\Rightarrow \lambda =2713\,\AA</math>

Revision as of 14:42, 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 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

The photoelectric equation then gives