Photoelectrons are emitted with maximum kinetic energy $E$ from a metal surface when light of frequency $ν$ falls on it. When light of frequency $ν'$ falls on the same metal, the maximum kinetic energy of emitted photoelectrons is found to be $2 E$ , then $ν'$ is

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$ν' = ν$

$ν' > 2 ν$

$ν' = 2 ν$

$ν' < 2 ν$

answer is C.

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Detailed Solution

Using photoelectricity equation

$E = h v + ϕ_{0} 2 E = h ν' + ϕ_{0} \Rightarrow 2 (h ν + ϕ_{0}) = h ν' + ϕ_{0}$

$\Rightarrow ν' = 2 ν + \frac{ϕ_{0}}{h} \Rightarrow ν' > 2 ν$

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