In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_{P}, E_{Q} a n d E_{R}$ , 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 metal 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 $4.0 e V, 4.5 e V$ and $5.5 e V$ , respectively. The energy of the incident photon used for metal $R$ , in $e V$ , is________

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answer is 6.

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Detailed Solution

For P and Q
Using Einstein’s P.E equation
$E - ϕ = h v \Rightarrow E_{1} - 4 = E_{P}$
Since, $E_{P} = 2 E_{Q}$
$∴ E_{1} - 4 = (E_{1} - 4.5) \Rightarrow E_{1} = 5 e V$
Thus, $E_{P} = 1 e V,$
$E_{Q} = E_{R} = 05 e V$
For $E_{2} - 5.5 = 0.5 \Rightarrow E_{2} = 6 e V$