The energy of a photon of wavelength \( 1\,\mathring{\mathrm{A}} \) is (Given: \( h = 6.63 \times 10^{-34}\,\mathrm{Js} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.24 \times 10^{4}\,\mathrm{eV} \)
The energy of a photon is given by the Planck relation.
\( \displaystyle E = h\nu = \frac{hc}{\lambda} \)
\( \lambda = 1\,\mathring{\mathrm{A}} = 10^{-10}\,\mathrm{m} \)
\( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle E = \frac{6.63\times10^{-34} \times 3\times10^{8}}{10^{-10}} = 1.989\times10^{-15}\,\mathrm{J} \)
Converting to electron volts with \( 1\,\mathrm{eV} = 1.6\times10^{-19}\,\mathrm{J} \):
\( \displaystyle E = \frac{1.989\times10^{-15}}{1.6\times10^{-19}}\,\mathrm{eV} \approx 1.24\times10^{4}\,\mathrm{eV} \)
Shortcut: Use \( hc \approx 12400\,\mathrm{eV}\,\mathring{\mathrm{A}} \) (that is, \( 1240\,\mathrm{eV\,nm} \)): \( E = 12400/1 = 1.24\times10^{4}\,\mathrm{eV} \).