In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are $E_{P} {, E}_{Q} {and E}_{R}$ , respectively, and they are related by $E_{P} = 2 E_{Q} {=2 E}_{R}$ . In this experiment, the same source of monochromatic light is used for metals P and Q while a different source of monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is:

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answer is 6.

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Detailed Solution

For Metal P, we have $E_{p} = \frac{h c}{λ_{1}} - 4$
For Metal Q, we have $E_{Q} = \frac{E_{p}}{2} = \frac{h c}{λ_{1}} - 4.5$
From equations (1) and (2), we have
$E_{p} = 1 e V$
For Metal R, we have $E_{R} = \frac{E_{p}}{2} = \frac{h c}{λ_{2}} - 5.5$
$\Rightarrow \frac{h c}{λ_{2}} = 6 e V$