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 ___.

For any metal, Einstein’s photoelectric equation is
$$h\nu - \phi = E_{\,\text{max}}$$
where $$h\nu$$ is the photon energy, $$\phi$$ is the work function and $$E_{\,\text{max}}$$ is the maximum kinetic energy of the emitted photoelectrons.

The three metals have maximum kinetic energies related by
$$E_P = 2E_Q = 2E_R \;\;\Longrightarrow\;\; E_Q = E_R$$ $$-(1)$$

Metals P and Q (same light source)
Let the common photon energy be $$h\nu_{PQ}$$.

For metal P:  $$h\nu_{PQ} - 4.0 = E_P$$ $$-(2)$$

For metal Q:  $$h\nu_{PQ} - 4.5 = E_Q$$ $$-(3)$$

Subtract $$(3)$$ from $$(2)$$:
$$(h\nu_{PQ} - 4.0) - (h\nu_{PQ} - 4.5) = E_P - E_Q$$
$$0.5 = E_P - E_Q$$ $$-(4)$$

But from $$E_P = 2E_Q$$ (from relation $$-(1)$$),
$$E_P - E_Q = 2E_Q - E_Q = E_Q$$

Comparing with $$-(4)$$:
$$E_Q = 0.5\;\text{eV}$$

Hence  $$E_P = 2E_Q = 1.0\;\text{eV}$$

Metal R (different light source)
Photon energy for this source is $$h\nu_R$$.

Using photoelectric equation for metal R:
$$h\nu_R - 5.5 = E_R$$

But $$E_R = E_Q = 0.5\;\text{eV}$$ (from $$-(1)$$).

Therefore  $$h\nu_R = 5.5 + 0.5 = 6.0\;\text{eV}$$

Hence, the energy of the incident photon used for metal R is 6 eV.