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For photoelectric emission from certain metal the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass):
A
$2\sqrt{h\nu/m}$
B
$\sqrt{h\nu/(2m)}$
C
$\sqrt{h\nu/m}$
D
$\sqrt{2h\nu/m}$
Detailed Solution
$h(2\nu) = h\nu + \frac{1}{2}mv_{max}^2$
$v_{max} = \sqrt{\frac{2h\nu}{m}}$
$v_{max} = \sqrt{\frac{2h\nu}{m}}$