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 $E_P=2E_Q=2E_R$. 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 $4.0\text{eV},4.5\text{eV}$ and $5.5\text{eV}$, respectively. The energy of the incident photon used for metal $R$, in $\text{eV}$, is

Answer 1

We know that,

$E=\frac{hc}{\lambda }-\varphi$

For $P$ & $Q$,

$E_P=\frac{hc}{{\lambda }_1}-4;E_Q=\frac{hc}{{\lambda }_1}-4.5$

$E_P=2E_Q$

From all above equations, we get

$E_Q=0.5\text{eV}$

$\therefore E_R=E_Q=0.5\text{eV}$

For metal $R$,

$E_R=\frac{hc}{{\lambda }_2}-5.5$

$\Rightarrow \frac{hc}{{\lambda }_2}=E_R+5.5=0.5+5.5=6\text{eV}$

Correct answer $:6$