The electric field of an electromagnetic wave in vacuum is given (in SI units) by \( E=E_0\sin\pi\left(4\times10^{6}x-1.2\times10^{15}t\right) \). The energy of a photon of this radiation is (Take \( h=6.63\times10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3.98\times10^{-19}\,\mathrm{J} \)
Comparing the field with \( E=E_0\sin(kx-\omega t) \), the coefficient of \( t \), with the factor \( \pi \) included, is the angular frequency. Each photon of radiation of frequency \( \nu \) carries energy \( E=h\nu \).
The angular frequency is:
\( \omega=1.2\pi\times10^{15}\,\mathrm{rad\,s^{-1}} \)
The frequency is:
\( \begin{aligned} \nu &= \frac{\omega}{2\pi} \\ &= \frac{1.2\pi\times10^{15}\,\mathrm{rad\,s^{-1}}}{2\pi} \\ &= 6\times10^{14}\,\mathrm{Hz} \end{aligned} \)
The energy of one photon is:
\( \begin{aligned} E &= h\nu \\ &= 6.63\times10^{-34}\,\mathrm{J\,s}\times6\times10^{14}\,\mathrm{Hz} \\ &\approx 3.98\times10^{-19}\,\mathrm{J} \end{aligned} \)