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| ''Source:'' "Theory and problems of Modern Physcis", Ronald Gautreau,Problem 9.13
| | The work function is |
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| '''Problem:''' The emitter in a photoelectric tube has a threshold wavelength of 6000 A. Determine the wavelength of the light incident on the tube if the stopping potential for this light is 2.5 V.
| | <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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| '''Solution:''' The work function is
| | The photoelectric equation then gives |
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| <math>eW_{0}=h\upsilon _{th}=\frac{hc}{\lambda }=\frac{12.4\times 10^{3 eV.A^{0}}}{6000A^{0}}=2.07 eV</math> | | <math>eV_{s}=h\nu-eW_{0}=\frac{hc}{\lambda}-eW_{0}</math> |
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| The photoelectric equation then gives
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| <math>ev_{s}=h\upsilon-eW_{0}=\frac{hc}{\lambda }-eW_{0}</math> | | <math>2.5\text{ eV}=\frac{12400 \text{ eV}\cdot\AA}{\lambda }-2.07\text{ eV}\Rightarrow \lambda =2713\,\AA</math> |
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| <math>2.5eV=\frac{12.4\times 10^{3}eV.A^{0}}{\lambda }-2.07eV\Rightarrow \lambda =2713A^{0}</math>
| | Back to [[Photoelectric Effect#Problem|Photoelectric Effect]] |
Latest revision as of 13:19, 18 January 2014
The work function is
The photoelectric equation then gives
Back to Photoelectric Effect