JEE Advanced 2021 Paper 2 · Q19 · Photoelectric Effect

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 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 _____.

SolveKar AI · Accepted Answer

Let $E_R = E$. Then $E_Q = E$ and $E_P = 2E$. Let photon energy for $P,Q$ be $h
u_{PQ}$ and for $R$ be $h
u_R$. Einstein photoelectric equation: $E_P = h
u_{PQ} - \phi_P \Rightarrow 2E = h
u_{PQ} - 4.0$ ... (i). $E_Q = h
u_{PQ} - \phi_Q \Rightarrow E = h
u_{PQ} - 4.5$ ... (ii). Subtract (ii) from (i): $E = 0.5$ eV. From (ii): $h
u_{PQ} = 5.0$ eV. For $R$: $E_R = h
u_R - \phi_R \Rightarrow 0.5 = h
u_R - 5.5 \Rightarrow h
u_R = 6.0$ eV. So photon energy for $R = 6$ eV.

JEE Advanced 2021 Paper 2 · Q19 · Photoelectric Effect

Solution