Power of the source, P=15 kW=15×103 J/s
Number of photons emitted per second, n=2.5×1022 s−1
Energy of one photon, E=nP=2.5×102215×103=6×10−19 J
Using E=λhc, we get λ=Ehc
λ=6×10−196.6×10−34×3×108=3.3×10−7 m=330 nm
The wavelength 330 nm lies in the ultraviolet region of the electromagnetic spectrum.
Answer: Ultraviolet