Question

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are EP, EQ and ER, respectively, and they are related by EP = 2EQ = 2ER. 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 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 ___.

Answer 1

$\begin{array}{l}\frac{hc}{\lambda _{1}}=\phi _{p}+E_{p}\end{array}$
$\begin{array}{l}\frac{hc}{\lambda _{1}}=\phi _{Q}+E_{Q}\end{array}$
Ep = 2EQ
EP– EQ = 0.5
⇒ Ep = 1.0 eV, EQ = 0.5 eV
ER = 0.5 eV
Energy of incident photon on R = φR + ER = 6 eV