Incident Photon Energy Calculation in Photoelectric Effect Experiment

According to Einstein's photoelectric equation, the maximum kinetic energy $E$ of ejected photoelectrons is given by: $E = h\nu - \Phi$ where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function of the metal.

Step 1: Determine the photon energy of the light source used for metals $P$ and $Q$

Let the energy of the monochromatic light source used for metals $P$ and $Q$ be $h\nu_1$.

For metal $P$: $E_P = h\nu_1 - \Phi_P = h\nu_1 - 4.0\text{ eV}$

For metal $Q$: $E_Q = h\nu_1 - \Phi_Q = h\nu_1 - 4.5\text{ eV}$

We are given that: $E_P = 2E_Q$

Substituting the expressions for $E_P$ and $E_Q$: $h\nu_1 - 4.0 = 2(h\nu_1 - 4.5)$ $h\nu_1 - 4.0 = 2h\nu_1 - 9.0$ $h\nu_1 = 5.0\text{ eV}$

Step 2: Calculate the maximum kinetic energy for metal $R$

Using the value of $h\nu_1$, we find: $E_P = 5.0\text{ eV} - 4.0\text{ eV} = 1.0\text{ eV}$

From the given relation $E_P = 2E_R$: $E_R = \frac{E_P}{2} = \frac{1.0\text{ eV}}{2} = 0.5\text{ eV}$

Step 3: Determine the energy of the incident photon used for metal $R$

Let the energy of the light source used for metal $R$ be $h\nu_2$.

For metal $R$: $E_R = h\nu_2 - \Phi_R$

Given $\Phi_R = 5.5\text{ eV}$ and $E_R = 0.5\text{ eV}$: $0.5\text{ eV} = h\nu_2 - 5.5\text{ eV}$ $h\nu_2 = 5.5\text{ eV} + 0.5\text{ eV} = 6.0\text{ eV}$

Thus, the energy of the incident photon used for metal $R$ is 6 $\text{eV}$ (or 6).