Question

Some satellites use nuclear power.

(a) If such a satellite emits a \({\rm{1}}{\rm{.00 - W}}\) flux of \(\gamma \) rays having an average energy of \({\rm{0}}{\rm{.500 MeV}}\), how many are emitted per second?

(b) These \(\gamma \) rays affect other satellites. How far away must another satellite be to only receive one \(\gamma \) ray per second per square meter?

Short answer

(a) The number of\(\gamma \)ray photons emitted per second are\(n = 1.25 \times {10^{13}}{\rm{ photons}}\;{{\rm{s}}^{ - 1}}\).

(b) Another satellite should be \(r = 997.4\;{\rm{km}}\) away to only receive one \(\gamma \) ray per second per square meter.

Step-by-step solution

Determine the formula for the energy and the 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 –

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

Consider the relation between joule and electron volt as:

\({\rm{1}}{\rm{.00}}\;{\rm{J}} = {\rm{6}}{\rm{.242}} \times {\rm{1}}{{\rm{0}}^{{\rm{18}}}}\;{\rm{eV}}\)

Consider the value of the energy is as follows:

\(\begin{array}{c}E = 0.500 \times {10^6}\;{\rm{eV}} \times \frac{{1.00\;\;{\rm{J}}}}{{6.242 \times {{10}^{18}}\;{\rm{eV}}}}\\ = 8.00 \times {10^{ - 13}}\;\;{\rm{J}}\end{array}\)

Determine the number of \(\gamma \) rays emitted

(a)

From equation (3) the number of \(\gamma \) ray photons emitted per second is obtained as:

\(\begin{array}{c}n = \frac{{1.00\;\;{\rm{W}}}}{{8.00 \times {{10}^{ - 13}}\;{\rm{J}}}}\\ = \frac{{1.00\;\;{\rm{J}}{{\rm{s}}^{ - {\rm{1}}}}}}{{8.00 \times {{10}^{ - 13}}\;{\rm{J}}}}\\ = 1.25 \times {10^{13}}{\rm{ photons}} \cdot {\rm{s}}{{\rm{ }}^{ - {\rm{1}}}}\end{array}\)

Therefore, the value for number of \(\gamma \) rays is \(n = 1.25 \times 1{{\rm{0}}^{{\rm{13}}}}{\rm{ photons}} \cdot {{\rm{s}}^{ - {\rm{1}}}}\).

Determine the distance of satellite

(b)

Now\(\gamma \)ray photon flux is defined as the number of\(\gamma \)ray photon per second per square meter and is given by –

\(\phi = \frac{n}{A}\) ….. (4)

Here, \(A = 4\pi {r^2}\) is the surface area of the satellite assuming shape of the satellite as spherical and \(r\) is the distance between the two satellites.

Therefore, from equation (4) rewrite the expression for the flux as:

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

Solving it for \(r\), it is obtained as:

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

From equation (5) \(\gamma \) ray photon flux received by the satellite

\(\phi = 1.00{\rm{ photon }}{{\rm{m}}^{ - 2}}\;{{\rm{s}}^{ - 1}}\)

Solve for the value of the radius as:

\(\begin{array}{c}r = \sqrt {\frac{{{\rm{1}}{\rm{.25}} \times {\rm{1}}{{\rm{0}}^{{\rm{13}}}}{\rm{ photons}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}{{{\rm{4}} \times \pi \times {\rm{1}}{\rm{.00 photo}}{{\rm{n}}^{ - {\rm{2}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}} \\ = \sqrt {\frac{{{\rm{1}}{\rm{.25}} \times {\rm{1}}{{\rm{0}}^{{\rm{13}}}}{\rm{ photons}} \cdot {\rm{s}}{{\rm{ }}^{{\rm{ - 1}}}}}}{{{\rm{4}} \times \pi \times {\rm{1}}{\rm{.00 photon }}{{\rm{m}}^{ - {\rm{2}}}} \cdot {{\rm{s}}^{{\rm{ - 1}}}}}}} \\ = {\rm{997356}}\;{\rm{m}}\\ = {\rm{997}}{\rm{.4\;}}\;{\rm{km}}\end{array}\)

Therefore, the value for distance is obtained as \(r = 997.4\;{\rm{km}}\).