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When photons of energy $h\nu$ fall on an aluminium plate (of work function $E_0$), photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be
A
$K + E_0$
B
$2K$
C
$K$
D
$K + h\nu$
Detailed Solution
Einstein's photoelectric equation: photon energy = work function + maximum kinetic energy.
In the first case: $h\nu = E_0 + K$ ...(1)
When the frequency is doubled, the photon energy becomes $2h\nu$: $2h\nu = E_0 + K'$ ...(2)
Subtracting (1) from (2): $h\nu = K' - K$
$\Rightarrow K' = K + h\nu$
Note that the kinetic energy is more than doubled, since $K' = 2K + E_0$.
In the first case: $h\nu = E_0 + K$ ...(1)
When the frequency is doubled, the photon energy becomes $2h\nu$: $2h\nu = E_0 + K'$ ...(2)
Subtracting (1) from (2): $h\nu = K' - K$
$\Rightarrow K' = K + h\nu$
Note that the kinetic energy is more than doubled, since $K' = 2K + E_0$.