The problem involves the photoelectric effect, which is described by Einstein's photoelectric equation: $K_{max} = h\nu - \phi$, where $K_{max}$ is the maximum kinetic energy of the photoelectrons, $h\nu$ is the energy of the incident photon, and $\phi$ is the work function of the metal.

Step 1: Set up the photoelectric equations for metals P and Q

Let the energy of the incident photons from the first source (used for metals P and Q) be $E_{ph1}$. The work functions are given as $\phi_P = 4.0$ eV and $\phi_Q = 4.5$ eV.

The photoelectric equations for metals P and Q are: For metal P: $E_P = E_{ph1} - \phi_P = E_{ph1} - 4.0$ (1) For metal Q: $E_Q = E_{ph1} - \phi_Q = E_{ph1} - 4.5$ (2)

Step 2: Use the given relation between kinetic energies to find the incident photon energy for P and Q

We are given the relation $E_P = 2E_Q$. Substituting the expressions from equations (1) and (2): $E_{ph1} - 4.0 = 2(E_{ph1} - 4.5)$ $E_{ph1} - 4.0 = 2E_{ph1} - 9.0$ $9.0 - 4.0 = 2E_{ph1} - E_{ph1}$ $E_{ph1} = 5.0 \text{ eV}$ So, the energy of the incident photons used for metals P and Q is 5.0 eV.

Step 3: Calculate the kinetic energies of the photoelectrons

Now we can calculate the maximum kinetic energies $E_P$ and $E_Q$: $E_P = E_{ph1} - 4.0 = 5.0 - 4.0 = 1.0$ eV $E_Q = E_{ph1} - 4.5 = 5.0 - 4.5 = 0.5$ eV (Check: $E_P = 1.0$ eV and $E_Q = 0.5$ eV, so $E_P = 2E_Q$ is satisfied.)

We are also given the relation $E_P = 2E_R$. Using the calculated value of $E_P$: $1.0 = 2E_R$ $E_R = \frac{1.0}{2} = 0.5 \text{ eV}$

Step 4: Find the energy of the incident photon for metal R

For metal R, a different source of monochromatic light is used. Let its photon energy be $E_{ph2}$. The work function for metal R is $\phi_R = 5.5$ eV. The photoelectric equation for metal R is: $E_R = E_{ph2} - \phi_R$ We know $E_R = 0.5$ eV and $\phi_R = 5.5$ eV. Substituting these values: $0.5 = E_{ph2} - 5.5$ $E_{ph2} = 0.5 + 5.5$ $E_{ph2} = 6.0 \text{ eV}$

Thus, the energy of the incident photon used for metal R is 6.0 eV.

The final answer is an integer value, which is 6.