To solve this problem, we will use the photoelectric equation:$\\ E = \phi + K.E. \\$where $E$ is the energy of the incident radiation, $\phi$ is the work function of the metal,$\\$and $K.E.$ is the kinetic energy of the emitted photoelectrons.$\\$Initially, let the energy of the incident radiation be $E_0. \\$The initial kinetic energy of the photoelectrons is $0.5$ eV.$\\$Using the photoelectric equation for the initial condition:$\\ E_0 = \phi + 0.5 \\$When the energy of the incident radiation is increased by $20\%, \\$the new energy $E_1$ becomes $1.2E_0. \\$The new kinetic energy of the photoelectrons is $0.8$ eV.$\\$Using the photoelectric equation for the new condition:$\\ 1.2E_0 = \phi + 0.8 \\$We now have two equations:$\\ 1. \quad E_0 = \phi + 0.5 \\ 2. \quad 1.2E_0 = \phi + 0.8 \\$Subtract equation 1 from equation 2:$\\ 1.2E_0 - E_0 = (\phi + 0.8) - (\phi + 0.5) \\ 0.2E_0 = 0.3 \\$Solve for $E_0: \\ E_0 = \frac{0.3}{0.2} = 1.5$ eV$\\$Substitute $E_0$ back into equation 1 to find $\phi: \\ 1.5 = \phi + 0.5 \\ \phi = 1.5 - 0.5 \\ \phi = 1.0$ eV$\\$Therefore, the work function of the metal is $1.0$ eV. This corresponds to Option 2.