Light of frequency $\nu = 2\nu_0$ ejects photoelectrons from a metal with threshold frequency $\nu_0$. The maximum KE of the photoelectrons is
A$2h\nu_0$
B$3h\nu_0$
C$h\nu_0$
D$h\nu_0 / 2$
Answer & Solution
Correct answer: C. $h\nu_0$
$KE_{max} = h\nu - h\nu_0 = h(2\nu_0) - h\nu_0 = h\nu_0$.
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