How much energy will be carried out by electromagnetic radiation of wavelength \( 3\times10^{-6}\,\mathrm{m} \)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 6.63\times10^{-20}\,\mathrm{J} \)
The energy carried by one photon of radiation of wavelength \( \lambda \) is given by Planck’s relation \( E = h\nu = hc/\lambda \).
\( \displaystyle E = \frac{6.63\times10^{-34}\times3\times10^{8}}{3\times10^{-6}} \)
\( E = 6.63\times10^{-20}\,\mathrm{J} \)
Shortcut: \( E(\mathrm{eV}) = 1240/\lambda(\mathrm{nm}) \). For \( 3000\,\mathrm{nm} \), \( E\approx0.41\,\mathrm{eV} \approx0.41\times1.6\times10^{-19}\,\mathrm{J}\approx6.6\times10^{-20}\,\mathrm{J} \), which immediately rules out options of the order \( 10^{47} \) or \( 10^{20} \).