(a) What is the ratio of power outputs by two microwave ovens having frequencies of \({\rm{950}}\) and \({\rm{2560 MHz}}\), if they emit the same number of photons per second?

(b) What is the ratio of photons per second if they have the same power output?

The energy of a photon having frequency \({\rm{f}}\) is –

\({\bf{E = hf}}\) …… (1)

Here, \(h = 6.626 \times {10^{ - 34}}\;{\rm{Js}}\), is Planck's constant, and \({\rm{f}}\) 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 –

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

Here given microwave frequencies are:

\(\begin{array}{c}{f_1} = 950\;{\rm{MHz}}\\ = 950 \times {10^6}\;{\rm{Hz}}\end{array}\)

Consider the second frequency is:

\(\begin{array}{}{f_2} &= 2560\;{\rm{MHz}}\\ &= 2560 \times {10^6}\;\;{\rm{Hz}}\end{array}\)

Photon emission time is –

\(t = 1.00\;\;{\rm{s}}\)

Now from equation\((1)\), the energy of the photon corresponding to the frequency \({{\rm{f}}_{\rm{1}}}\) is:

\({E_1} = h{f_1}\)

Substitute the values and solve as:

\(\begin{array}{}{E_1} &= 6.626 \times {10^{ - 34}}\;{\rm{Js}} \times 950 \times {10^6}\;\;{\rm{Hz}}\\ &= 6.30 \times {10^{ - 25}}\;\;{\rm{J}}\end{array}\)

And the energy of the photon corresponding to the frequency \({{\rm{f}}_{\rm{2}}}\) is:

\({E_2} = h{f_2}\)

Substitute the values and solve as:

\(\begin{array}{}{E_2} &= 6.626 \times {10^{ - 34}}\;{\rm{Js}} \times 2560 \times {10^6}\;\;{\rm{Hz}}\\ & = 1.70 \times {10^{ - 24}}\;{\rm{J}}\end{array}\)

Now, from equation\((2)\), the power of the \({{\rm{n}}_{\rm{1}}}\) photons emitted per second corresponding to the energy \({{\rm{E}}_{\rm{1}}}\) is:

\({P_1} = {n_1}\frac{{{E_1}}}{t}\)

Substitute the values and solve as:

\(\begin{array}{}{P_1} &= {n_1}\frac{{6.30 \times {{10}^{ - 25}}\;{\rm{J}}}}{{1.00\;\;{\rm{s}}}}\\ &= \left( {6.30 \times {{10}^{ - 25}}\;\;{\rm{W}}} \right){n_1}\end{array}\)

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

\({P_2} = {n_2}\frac{{{E_2}}}{t}\)

Solve further as:

\(\begin{array}{}{P_2} = {n_2}\frac{{{\rm{1}}{\rm{.70}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{24}}}}\;{\rm{J}}}}{{{\rm{1}}{\rm{.00\;s}}}}\\{\rm{ = }}\left( {{\rm{1}}{\rm{.70 \times 1}}{{\rm{0}}^{ - {\rm{24}}}}\;{\rm{W}}} \right){{\rm{n}}_{\rm{2}}}\end{array}\)

(a)

Now from equations(3)and(4), the ratio of power output corresponding to the frequencies\({{\rm{f}}_{\rm{1}}}\)and\({{\rm{f}}_{\rm{2}}}\)is as follows:

\({R_P} = \frac{{{P_1}}}{{{P_2}}}\) …… (5)

From the question:

\({n_1} = {n_2}\) ……. (6)

From equations (5) and (6).

\(\begin{array}{}{R_P} &= \frac{{{P_1}}}{{{P_2}}}\\ & = \frac{{\left( {6.30 \times {{10}^{ - 25\;{\rm{W}}}}} \right){n_2}}}{{\left( {1.70 \times {{10}^{ - 24\;{\rm{W}}}}} \right){n_2}}}\\ &= 0.371\end{array}\)

Therefore, the value for ratio is obtained as \({R_P} = 0.371\).

(b)

Now from equation (5), ratio of photons per second is –

\({R_n} = \frac{{{P_1}}}{{{P_2}}}\) …… (7)

Now according to the question if –

\({P_1} = {P_2}\)

Then from equation (5) solve as:

\(\begin{array}{l}0.371 \times \frac{{{n_1}}}{{{n_2}}} = 1\\\frac{{{n_1}}}{{{n_2}}} = \frac{1}{{0.371}}\\\frac{{{n_1}}}{{{n_2}}} = 2.70\end{array}\) …… (8)

Hence, from equations (7) and (8) the value of the ratio is:

\({R_n} = 2.70\)

Therefore, the value for ratio is obtained as \({R_n} = 2.70\).