Question

(a) How far away must you be from a \({\bf{650 - kHz}}\) radio station with power \({\bf{50}}{\bf{.0 kW}}\) for there to be only one photon per second per square meter? Assume no reflections or absorption, as if you were in deep outer space.

(b) Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts.

Short answer

(a) One must be\(r = 0.321\;{\rm{ly}}\)away from a\({\rm{650 kHz}}\)radio station with power\({\rm{50}}{\rm{.0 kW}}\)for there to be only one photon per second per square meter.

(b) The signals from other solar systems cannot be detected as the photon flux and transmitted power is very low.

Step-by-step solution

Determine the formula for the energy, power and number of photons.

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

The power is defined as the energy per second is obtained as:

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

Therefore, the number of emitted photons per second is obtained as:

\({\bf{n = }}\frac{{\bf{P}}}{{\bf{E}}}\) …… (3)

Consider the expression for the flux as follows:

\(\phi {\bf{ = }}\frac{{\bf{n}}}{{\bf{A}}}\) …… (4)

Here, \(A = 4\pi {r^2}\) is the surface area of the receiving object assuming shape of the object as spherical and \(r\) is the distance between the radio station and receiving object.

From equation (1), write the value of the energy in Joule as follows:

\(\begin{align}{}E &= hf\\ & = 6.626 \times {10^{ - 34}}\;{\rm{Js}} \times 650 \times {10^3}{\rm{\;Hz}}\\ & = 4.31 \times {10^{ - 28}}\;{\rm{J}}\end{align}\)

Determine the number of photons

(a)

From equation\({\rm{(3)}}\), the number of emitted photons per second is obtained as:

\(\begin{align}{}n & = \frac{{50.0 \times {{10}^3}\;{\rm{W}}}}{{4.31 \times {{10}^{ - 28}}\;{\rm{J}}}}\\& = \frac{{50.0 \times {{10}^3}\;\;{\rm{J\;}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}{{4.31 \times {{10}^{ - 28}}\;{\rm{J}}}}\\ & = 1.16 \times {10^{32}}\;{\rm{photons}} \cdot {\rm{s}}{{\rm{ }}^{ - {\rm{1}}}}\end{align}\)

Therefore, from equation\((4)\), it is obtained –

\(\phi = \frac{n}{{4\pi {r^2}}}\) ….. (5)

Rewrite the expression for \(r\) as follows:

\(r = \sqrt {\frac{n}{{4\pi \phi }}} \) ……. (6)

Here given that the photon flux received by the receiving object is as follows:\(\phi = 1.00{\rm{ photon}}\;{{\rm{m}}^{ - 2}} \cdot {{\rm{s}}^{ - 1}}\)

Solve for the radius as:

\(\begin{array}{c}r = \sqrt {\frac{{{\rm{1}}{\rm{.16}} \times {\rm{1}}{{\rm{0}}^{{\rm{32}}}}{\rm{ photon}}{{\rm{s}}^{ - {\rm{1}}}}}}{{{\rm{4}} \times \pi \times {\rm{1}}{\rm{.00 photo}}{{\rm{n}}^{ - {\rm{2}}}}{\rm{\;}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}} \\ = {\rm{3}}{\rm{.04}} \times {\rm{1}}{{\rm{0}}^{{\rm{15}}}}\;{\rm{m}}\end{array}\)

Since, \({\rm{1}}{\rm{.00}}\;{\rm{ly}} = {\rm{9}}{\rm{.461}} \times {\rm{1}}{{\rm{0}}^{{\rm{15}}}}\;{\rm{m}}\)

Solve further as,

\(\begin{array}{c}r = 3.04 \times {10^{15}}\;{\rm{m}} \times \frac{{1.00\;{\rm{ly}}}}{{9.461 \times {{10}^{15}}\;{\rm{\;m}}}}\\ = 0.321\;\;{\rm{m}}\end{array}\)

Therefore, the value for distance is obtained as \(r = 0.321\;{\rm{ly}}\).

Implications for detecting intelligent life

(b)

The distance between the radio station and receiving object is a distance where the photon flux becomes \(\phi = 1.00{\rm{ photon }}{{\rm{m}}^{ - 2}}\;{{\rm{s}}^{ - 1}}\). But the distance between our solar system and other solar system is very large than \(r = 3.04 \times {10^{15}}\;m = 0.321\;ly\).

Therefore, to determine the presence of intelligent life in other solar systems by with the help of radio broadcasts, the value of photon flux must be higher at the receiver end. Therefore, the transmitted power from the other solar system should be tremendously huge for the photon flux to be sufficient to reach the other solar system.