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What is the stopping potential when the metal with work function `0.6 eV` is illuminated with the light of `2 eV` ?A. `2.6 V`B. `3.6 V`C. `0.8 V`D. `1.4 V`

Answer» Correct Answer - D
`V_(0) = ((E - W_(0)))/ ( e ) = (( 2eV - 0.6 eV))/( e ) = 1.4 V`


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