Use Einstein’s photoelectric equation

The maximum kinetic energy is $K_{\max} = E_{\text{incident}} - \phi$ where $\phi$ is the work function.

Find the work function using threshold wavelength

Given threshold wavelength: $\lambda_0 = 380\,\text{nm}$

Using $E(\text{in eV}) = \frac{1237}{\lambda(\text{in nm})}$

so the work function is $\phi = \frac{1237}{380} \approx 3.255\,\text{eV}$

Find the energy of incident photon

Given incident wavelength: $\lambda = 260\,\text{nm}$

Therefore, $E_{\text{incident}} = \frac{1237}{260} \approx 4.758\,\text{eV}$

Calculate maximum kinetic energy

$K_{\max} = 4.758 - 3.255 = 1.503\,\text{eV}$

$K_{\max} \approx 1.5\,\text{eV}$

Match with the options

The correct option is: $\boxed{\text{D: }1.5\,\text{eV}}$