91Ó°ÊÓ

From the equation, the energy of a photon is given by

\(E = hf\) ...(1)

Where \(h = 6.626 \times {10^{ - 34}}\,{\rm{Js}}\) is Planck's constant,

\(f\) is the frequency of the incident photon.

Here the frequency of the photon in a radio wave from an AM station is

\(\begin{array}{c}f = 1530\,{\rm{kHz}}\\ = 1530 \times {10^3}\,{\rm{Hz}}\end{array}\)

Therefore, from equation, we get

\(E = hf\)

Substitute all the value in the above equation

\(\begin{array}{c}E = 6.626 \times {10^{ - 34}}\,{\rm{Js}} \times 1530 \times {10^3}\,{\rm{Hz}}\\ &= 1.01 \times {10^{ - 27}}\,{\rm{J}}\end{array}\)

Since, we know

\(1.00\;\,{\rm{J}} = 6.242 \times {10^{18}}\,{\rm{eV}}\)

Therefore we get

\(\begin{array}{c}E = 1.01 \times {10^{ - 27}}\,{\rm{J}} \times \frac{{6.242 \times {{10}^{18}}\,{\rm{eV}}}}{{1.00\,{\rm{J}}}}\\ = 6.33 \times {10^{ - 9}}\,{\rm{eV}}\end{array}\)

Hence the energy in joules and eV of a photon are, \(1.01 \times {10^{ - 27}}\,{\rm{J}}\], \[6.33 \times {10^{ - 9}}\,{\rm{eV}}\).