$$ is the kinetic energy of the ejected electrons,

$$ E = h\nu $$ is the energy of the incident photon,

$$ h $$ is Planck's constant,

$$ \nu $$ is the frequency of the incident photon,

$$ W = h\nu_0 $$ is the work function of the material, and

$$ \nu_0 $$ is the threshold frequency.

To find the kinetic energy in electron-volts (eV), we first need to calculate the energy of the incident photons and the work function using the provided values. The energy of the incident photon (E) is given to be $$10 \mathrm{~eV}$$ directly.

The work function (W), however, needs to be calculated. It is the energy corresponding to the threshold frequency ($$\nu_0$$), so we can calculate it using:

$$ W = h\nu_0 $$

Given that $$h = 6.63 \times 10^{-34} \mathrm{~Js}$$ (or meter2 kg / s in SI units), and $$\nu_0 = 2 \times 10^{15} \mathrm{~Hz}$$, let's calculate W. However, since our final answer is desired in electron-volts, we must remember that $$1 \mathrm{~eV} = 1.602 \times 10^{-19} \mathrm{~J}$$.

To calculate the work function in eV, we use:

$$ W = h\nu_0 = (6.63 \times 10^{-34} \mathrm{~Js}) \times (2 \times 10^{15} \mathrm{~Hz}) $$

$$ W = 1.326 \times 10^{-18} \mathrm{~J} $$

Convert W to eV:

$$ W = \frac{1.326 \times 10^{-18} \mathrm{~J}}{1.602 \times 10^{-19} \mathrm{~eV/J}} = 8.29 \mathrm{~eV} $$

Now, we can calculate the kinetic energy of the photoelectrons:

$$ KE = E - W = 10 \mathrm{~eV} - 8.29 \mathrm{~eV} $$

$$ KE = 1.71 \mathrm{~eV} $$

Therefore, the kinetic energy of the photoelectrons emitted is 1.71 eV, which corresponds to Option D.

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