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(D) The power $P$ is given by $P = n E_{photon} = n \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.Rearranging the formula

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Ultraviolet

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(D) The power $P$ is given by $P = n E_{photon} = n \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.Rearranging the formula to solve for wavelength $\lambda$,we get $\lambda = \frac{nhc}{P}$.Substituting the given values: $n = 2.5 	imes 10^{22} 	ext{ s}^{-1}$,$h = 6.6 	imes 10^{-34} 	ext{ J}\cdot	ext{s}$,$c = 3 	imes 10^8 	ext{ m/s}$,and $P = 15 	ext{ kW} = 15 	imes 10^3 	ext{ W}$.$\lambda = \frac{(2.5 	imes 10^{22}) 	imes (6.6 	imes 10^{-34}) 	imes (3 	imes 10^8)}{15 	imes 10^3} = \frac{49.5 	imes 10^{-4}}{15 	imes 10^3} = 3.3 	imes 10^{-7} 	ext{ m}$.Converting to nanometers: $3.3 	imes 10^{-7} 	ext{ m} = 330 	ext{ nm}$.The wavelength range for the ultraviolet region is approximately $10 	ext{ nm}$ to $400 	ext{ nm}$.Therefore,$330 	ext{ nm}$ falls in the Ultraviolet region.

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