The energy of a photon of wavelength \(400\,\mu\mathrm{m}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3.09\times10^{-3}\,\mathrm{eV}\)
The energy of a photon is given by Planck’s relation \( E = h\nu = hc/\lambda \).
\( \displaystyle \lambda = 400\,\mu\mathrm{m} = 400\times10^{-6}\,\mathrm{m} = 4\times10^{-4}\,\mathrm{m} \)
\( \displaystyle E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34}\,\mathrm{J\,s}) \times (3\times10^{8}\,\mathrm{m\,s^{-1}})}{4\times10^{-4}\,\mathrm{m}} \)
\( \displaystyle E \approx 4.97\times10^{-22}\,\mathrm{J} \)
\( \displaystyle E = \frac{4.97\times10^{-22}}{1.6\times10^{-19}}\,\mathrm{eV} \approx 3.1\times10^{-3}\,\mathrm{eV} \)
This is the value given in Option (4), \( 3.09\times10^{-3} \) eV, to the rounding of the constants used.
Shortcut: With \( E(\mathrm{eV}) = \frac{1240}{\lambda(\mathrm{nm})} \) and \( 400\,\mu\mathrm{m} = 4\times10^{5}\,\mathrm{nm} \): \( E = 1240/(4\times10^{5}) \approx 3.1\times10^{-3} \) eV.