In an atom, the frequency of the radiation for a transition between two energy levels is \( 4\times10^{15}\,\mathrm{Hz} \). What is the corresponding energy? (Take \( h=6.6\times10^{-34}\,\mathrm{J\,s} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 16.5\,\mathrm{eV} \)
A photon of frequency \( \nu \) carries the energy difference in a transition between two energy levels. The energy of the photon is \( E=h\nu \). The result in joules is then changed to electron volts.
\( \begin{aligned} E &= h\nu \\ &= 6.6\times10^{-34}\,\mathrm{J\,s}\times4\times10^{15}\,\mathrm{Hz} \\ &= 2.64\times10^{-18}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} E &= \frac{2.64\times10^{-18}\,\mathrm{J}}{1.6\times10^{-19}\,\mathrm{J/eV}} \\ &= 16.5\,\mathrm{eV} \end{aligned} \)