When photons of energy h fall on an aluminium plate ( of work function ), photoelectrons of maximum kinetic energy of K are ejected. If the frequency of the radiation is doubled then ejected photoelectrons will be.

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K + hv

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Solution

The correct option is C K + hv
Maximum kinetic energy of photo electrons $K = E_{i} - ϕ$
where $ϕ$ is the work function of aluminum.
Case 1 :
Incident energy of photon $E_{i} = h v$
$∴$ $K = h v - ϕ$
We get $ϕ = h v - K$
Case 2 :
Frequency of the incident photon is doubled, so energy of incident photon also gets doubled.
Thus incident energy of photon $E_{i} = 2 h v$
$∴$ $K^{'} = 2 h v - ϕ$
$⟹ K^{'} = 2 h v - (h v - K) = K + h v$

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