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  • Photoelectron are emitted with maximum kinetic energy  from a metal surface when light of frequency <img alt="\upsilon" s

Photoelectron are emitted with maximum kinetic energy E from a metal surface when light of frequency \upsilon falls on it when light of frequency \upsilon' falls on the same metal, the max. KE . Of emitted Photoelectrons is found to be 2E then \upsilon' is -

Option: 1

\upsilon ^{'} = \upsilon


Option: 2

\upsilon^{'}= 2\upsilon


Option: 3

\upsilon^{'}> 2\upsilon


Option: 4

\upsilon^{'}< 2\upsilon


Answers (1)

best_answer

\mathrm{KE}=\mathrm{h\upsilon}+\phi                 ................(i)
2KE= h\upsilon '+\phi            ................\left ( ii \right )
or  2(h v+\phi)=h v^{\prime}+\phi
or v^{\prime}=2 v+\frac{\phi}{h} \Rightarrow v^{\prime}>2 v

Posted by

Irshad Anwar

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