The photoeletric threshold 4v is incident on the metal is v. When light of freqency 4v is incident on the metal, the maximum kinetic energy of the emitted photoelectron is
A. 4hv
B. 3hv
C. 5hv
D. `(5hv)/(2)`

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1 Answers

Correct Answer - B
(b) : The Maximum kinetic energy of the emitted electron is given by
`K_(max)=hv-phi_(0)=h(4v)-h(v)=3hv`

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