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  • In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with lithium cathode having work function . If the wavelength of

In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with lithium cathode having work function \phi = 2\cdot 5eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. \left ( h= 6\cdot 63\times 10^{-34}Js,c= 3\times 10^{8}ms^{-1} \right )
Option: 1 1\cdot 1V
Option: 2 0\cdot 6V
Option: 3 1\cdot 3V
Option: 4 1\cdot 9V

Answers (1)

best_answer

\lambda_{1}= 280\, nm
\phi_{0}= 2\cdot 5\, ev
\lambda_{2}= 400\, nm
By Einstein's photoelectric eqn,
\frac{hc}{\lambda_{1}}= \phi_{0}+e\left ( v_{0} \right )_{1}
E\left ( eV \right )= \frac{1240}{\lambda\left ( nm \right )}
\left ( \frac{hc}{\lambda_{1}} \right )\left ( eV \right )= \frac{1240}{280}= \frac{31}{7}
\left ( \frac{hc}{\lambda_{1}} \right )\left ( eV \right )= 4\cdot 428\, eV
\left ( \frac{hc}{\lambda_{2}} \right )eV= \frac{1240}{400}= 3\cdot 1\, eV
\frac{hc}{\lambda_{1}}-\frac{hc}{\lambda_{2}}= \left ( \phi_{0}-\phi_{0} \right )+e\left ( v_{0} \right )_{1}-e\left ( v_{0} \right )_{2}
1\cdot 328\, eV= e\left [\left ( v_{0} \right )_{1} -\left ( v_{0} \right )_{2} \right ]
\Delta v_{0}= 1\cdot 328\, V
The correct option is (3)
 

Posted by

vishal kumar

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