Question

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 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 $4.0\, eV , 4.5\, eV$ and $5.5 \,eV$, respectively. The energy of the incident photon used for metal $R$, in $e V$, is ______.

Answer 1

Let $E _{ R }= E$
Then $E _{ Q }= E$
$E _{ P }=2 E$
From Einstein's equation for $P , Q$ and $R$
$2 E = hv -4.0 \ldots$ (i)
$E = hv -4.5 \ldots$ (ii)
$E = hv ^{\prime}-5.5 $...(iii)
Solving equation (i) and (ii) we get
$E =0.5$
So, $hv ^{\prime}=6.0 \,eV$