Electromagnetic Radiation Spectrum Region for Monochromatic Source

To determine the region of the electromagnetic spectrum to which the emitted radiation belongs, we can calculate the wavelength of the emitted photons using the source's power and photon emission rate.

Step 1: Given Data

• Power of the source, $P = 15\text{ kW} = 15 \times 10^3\text{ J/s}$
• Rate of photon emission, $n = 2.5 \times 10^{22}\text{ photons/s}$
• Planck's constant, $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$
• Speed of light, $c = 3 \times 10^8\text{ m/s}$

Step 2: Calculate the energy of a single photon The power emitted by the source is equal to the energy of one photon ($E$) multiplied by the number of photons emitted per second ($n$): $P = n \cdot E$

Rearranging for $E$: $E = \frac{P}{n} = \frac{15 \times 10^3\text{ J/s}}{2.5 \times 10^{22}\text{ s}^{-1}} = 6 \times 10^{-19}\text{ J}$

Step 3: Calculate the wavelength ($\lambda$) of the photon Using the relation $E = \frac{hc}{\lambda}$: $\lambda = \frac{hc}{E}$

Substitute the given values into the equation: $\lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}}$

$\lambda = \frac{19.8 \times 10^{-26}}{6 \times 10^{-19}} = 3.3 \times 10^{-7}\text{ m} = 330\text{ nm}$

Step 4: Identify the region of the electromagnetic spectrum

• The visible spectrum ranges approximately from $400\text{ nm}$ to $700\text{ nm}$.
• Radiation with a wavelength of $330\text{ nm}$ lies just below the visible spectrum ($10\text{ nm}$ to $400\text{ nm}$), which corresponds to the Ultraviolet region.

Hence, the correct option is D.