Question

What is the energy in joules and eV of a photon in a radio wave from an AM station that has a \({\rm{1530 - kHz}}\) broadcast frequency?

Short answer

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

Step-by-step solution

Frequency of the photon in a radio wave

From 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

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

Energy in joules

Therefore, from equation, we get

\(E = hf\)

Substitute all the value in the above equation

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

Since, we know

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

Therefore we get

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