If a source of electromagnetic radiation having power 15 kW produces $$10^{16}$$ photons per second, the radiation belongs to a part of spectrum is:
(Take Planck constant $$h = 6 \times 10^{-34}$$ J s)

The source delivers power $$P = 15 \text{ kW} = 15 \times 10^{3}\, \text{J s}^{-1}$$.

Number of photons emitted each second $$n = 10^{16}\, \text{s}^{-1}$$.

Energy carried by one photon is obtained by dividing the total energy per second by the number of photons per second:

$$E_{\text{photon}} = \frac{P}{n} = \frac{15 \times 10^{3}}{10^{16}}$$

Simplifying,

$$E_{\text{photon}} = 1.5 \times 10^{-12}\, \text{J}$$

Planck’s relation connects the energy of a photon with its frequency: $$E_{\text{photon}} = h \nu$$, where $$h = 6 \times 10^{-34}\, \text{J s}$$.

Thus,

$$\nu = \frac{E_{\text{photon}}}{h} = \frac{1.5 \times 10^{-12}}{6 \times 10^{-34}}$$

$$\nu = 0.25 \times 10^{22}\, \text{Hz} = 2.5 \times 10^{21}\, \text{Hz}$$

Frequencies around $$10^{19}\, \text{Hz}$$ and higher correspond to the $$\gamma$$-ray region of the electromagnetic spectrum. Since $$2.5 \times 10^{21}\, \text{Hz}$$ lies well within this range, the radiation is classified as gamma rays.

Therefore, the correct option is Gamma rays (Option C).