The incident photons in a photo-electric effect experiment have wavelength of \( 7\times10^{-7} \) m. The energy of the photon will be:
(Given: h = \( 6.6\times10^{-34} \) Js)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.83\times10^{-19} \) J
The energy of a photon is \( E = h\nu = hc/\lambda \), with \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \).
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\,\mathrm{J\,s} \times 3\times10^{8}\,\mathrm{m\,s^{-1}}}{7\times10^{-7}\,\mathrm{m}} \)
\( \displaystyle E = \frac{19.8\times10^{-26}}{7\times10^{-7}}\,\mathrm{J} \approx 2.83\times10^{-19}\,\mathrm{J} \)
Shortcut: With \( E(\mathrm{eV}) \approx 1240/\lambda(\mathrm{nm}) \): \( \lambda = 700 \) nm gives about 1.77 eV, and \( 1.77 \times 1.6\times10^{-19} \approx 2.83\times10^{-19} \) J.