A radio transmitter radiates 1 kW of power at a wavelength of 198.6 m. How many photons does it emit per second? (Take \( h=6.62\times10^{-34}\,\mathrm{J\,s} \) and \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 10^{30} \)
Radiation of wavelength \( \lambda \) is emitted as photons, each of energy \( E=h\nu \), that is, \( hc/\lambda \). The number of photons emitted per second is the power divided by the energy of one photon.
The energy of each photon is:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{6.62\times10^{-34}\,\mathrm{J\,s}\times3\times10^{8}\,\mathrm{m\,s^{-1}}}{198.6\,\mathrm{m}} \\ &= 1\times10^{-27}\,\mathrm{J} \end{aligned} \)
The number emitted per second is:
\( \begin{aligned} n &= \frac{P}{E} \\ &= \frac{10^{3}\,\mathrm{J\,s^{-1}}}{1\times10^{-27}\,\mathrm{J}} \\ &= 10^{30}\,\mathrm{s^{-1}} \end{aligned} \)