Photoelectric Example: Difference between revisions

Revision as of 13:18, 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

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

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

Revision as of 15:49, 11 April 2013 (view source) RobertThrockmorton (talk \| contribs) No edit summary ← Older edit		Revision as of 13:18, 18 January 2014 (view source) RobertThrockmorton (talk \| contribs) No edit summary Newer edit →
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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>		<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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