In a photoemission experiment, the maximum kinetic energies of...

Let the maximum kinetic energies of photoelectrons be${E}_{P} = 2{E}_{Q} = 2{E}_{R} = 2E$.

The energy of incident photon on metal $P\text{and}Q$ be $U and for$R$be$U '$.</p><p>Photoelectric equation,${E}_{ph} = ϕ + K{E}_\max$</p><p>Now, using it for the three cases, </p><p>$U = 4 + 2E . . . \left(1\right)U = 4 . 5 + E . . . \left(2\right)U ' = 5 . 5 + E . . . \left(3\right)

From \left(1\right) \text{and}\left(2\right)$,$E = 0 . 5eV$</p><p>From \left(3\right)$, energy of the incident photon for metal $R$,

$U ' = 5 . 5 + 0 . 5 = 6eV$