A $$10$$ kW transmitter emits radio waves of wavelength $$500$$ m. The number of photons emitted per second by the transmitter is of the order of

The average rate at which energy is being radiated is the transmitter power $$P = 10 \text{ kW} = 10^{4}\, \text{J s}^{-1}$$.

Energy of one photon of wavelength $$\lambda$$ is given by the Planck relation $$E = \frac{hc}{\lambda}$$ where $$h = 6.626 \times 10^{-34}\, \text{J s}$$ and $$c = 3.0 \times 10^{8}\, \text{m s}^{-1}$$.

Therefore, the number of photons emitted each second is $$n = \frac{\text{power}}{\text{energy per photon}} = \frac{P}{E} = \frac{P\,\lambda}{h\,c}$$.

Substituting the given values: $$n = \frac{10^{4}\;\text{J s}^{-1} \times 500\;\text{m}}{6.626 \times 10^{-34}\;\text{J s} \times 3.0 \times 10^{8}\;\text{m s}^{-1}}$$ $$= \frac{5.0 \times 10^{6}}{1.988 \times 10^{-25}}$$ $$\approx 2.5 \times 10^{31}$$.

The order of magnitude is $$10^{31}$$.

Option B which is: $$10^{31}$$