Atomic StructureEasy
Question
Radiations of frequency, (ν) are incident on a photosensitive metal. The maximum kinetic energy of photoelectrons is E. When the frequency of the incident radiations is doubled, then what is the maximum kinetic energy of the photoelectrons?
Options
A.2E
B.E/2
C.E + hv
D.E – hv
Solution
${(K.E.)\text{ and }}_{\max}$
$\therefore E' = E + h\upsilon$
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