Question

(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?

Short answer

(a) The ratio of power outputs by two microwave ovens having frequencies of\({\rm{950 - 2560 MHz}}\), if they emit the same number of photons per second is,\({R_P} = 0.371\).

(b) The ratio of photons per second if they have the same power output is, \({R_n} = 2.70\).

Step-by-step solution

Determine the equation for energy and power,

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:

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

Determine the Ratio Calculation

(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 \({R_P} = 0.371\).

Determine the Ratio of Photons

(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 \({R_n} = 2.70\).