A) \[\sqrt{\frac{2h}{m}\left( {{\lambda }_{o}}-\lambda \right)}\]
B) \[\sqrt{\frac{2hc}{m}\left( {{\lambda }_{o}}-\lambda \right)}\]
C) \[\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{o}}-\lambda }{\lambda {{\lambda }_{o}}} \right)}\]
D) \[\sqrt{\frac{2h}{m}\left( \frac{1}{{{\lambda }_{o}}}-\frac{1}{\lambda } \right)}\]
Correct Answer: C
Solution :
The kinetic energy of the ejected electron is given by the equation \[hv=h{{v}_{0}}+\frac{1}{2}m{{v}^{2}}\because v=\frac{c}{\lambda }\]or \[\frac{hc}{\lambda }=\frac{hc}{{{\lambda }_{0}}}+\frac{1}{2}m{{v}^{2}}\] \[\frac{1}{2}m{{v}^{2}}=\frac{hc}{\lambda }=-\frac{hc}{{{\lambda }_{0}}}\] \[=hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\] \[\therefore \]\[{{v}^{2}}=\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\]or\[v=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)}\]You need to login to perform this action.
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