How many photons per second are emitted by the \({\bf{1}}{\bf{.00 kW}}\)antenna of a microwave oven, if its power output is at a frequency of \({\bf{2560 MHz}}\)?

The energy of a photon having frequency \({\rm{f}}\) is as follows:

\({\bf{E = hf}}\) …… (1)

Here, \(h = 6.626 \times {10^{ - 34}}\;{\rm{Js}}\), is Planck's constant, and \(f\) is the frequency of the incident photon.

The power is defined as the energy per second and given by –

\({\bf{P = }}\frac{{\bf{E}}}{{\bf{t}}}\) ...... (2)

Therefore, the number of emitted photons per second is given by –

\({\bf{n = }}\frac{{\bf{P}}}{{\bf{E}}}\)

From equation \({\rm{(1)}}\) the energy of the photon corresponding to the frequency \({\rm{f}}\) is as follows:

\(\begin{array}{c}E = 6.626 \times {10^{ - 34}}\;{\rm{Js}} \times 2560 \times {10^6}\;{\rm{Hz}}\\ = 1.70 \times {10^{ - 24}}\;\;{\rm{J}}\end{array}\)

From equation (2), the power of the \({\rm{n}}\) photons emitted per second corresponding to the energy \({\rm{E}}\) is as follows:

\(P = n\frac{E}{t}\) …… (3)

Substitute the values and solve as:

\(\begin{array}{l}1.00 \times {10^3}\;\;{\rm{W}} = n \times \frac{{1.70 \times {{10}^{ - 24}}\;\;{\rm{J}}}}{{1.00\;{\rm{s}}}}\\n = 5.88 \times {10^{26}}\;{\rm{photons}}\end{array}\)

Therefore, the value for number of photons is obtained as \(n = 5.88 \times {10^{26}}\) photons.