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Photoelectric emission is observed from a metal surface with incident frequencies \(\nu\)1 and \(\nu\)2 where \(\nu\)1 > \(\nu\)2 If the kinetic energies of the photoelectrons emitted in the two cases are in the ration 2:1, then the threshold frequency \(\nu\)o of the metal isa.\(\nu\)1 - \(\nu\)2b. \(\frac{\nu_!-\nu_2}{h}\)c. 2\(\nu\)1 - \(\nu\)2d. 2\(\nu\)2 - \(\nu\)1 |
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Answer» Correct option is d. 2\(\nu\)2 - \(\nu\)1 |
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