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Chapter 8: Problem 102

Radio signals from Voyager 1 in the 1970 s were broadcast at a frequency of 8.4 GHz. On Earth, this radiation was received by an antenna able to detect signals as weak as \(4 \times 10^{-21} \mathrm{W}\). How many photons per second does this detection limit represent?

Short Answer

Expert verified
The detection limit represents approximately \(7.18 \times 10^{2}\) photons per second.

Step by step solution

01

Determine the frequency in Hz

First, we need to convert the frequency given in GHz to Hz, as the standard unit for frequency in the photon energy formula is Hz. As 1 GHz = \(1 \times 10^{9}\) Hz, this implies that 8.4 GHz = \(8.4 \times 10^{9}\) Hz.
02

Calculate the energy of a single photon

Next, we're going to compute the energy of a single photon by using Planck's equation, \(E = h \cdot f\), where \(E\) represents the photon's energy, \(f\) is its frequency, and \(h\) denotes the Planck constant (\(6.63 \times 10^{-34}\) joules-second). Substituting the given frequency, the energy of one photon becomes \(E = 6.63 \times 10^{-34} \cdot 8.4 \times 10^{9} = 5.57 \times 10^{-24} \) joules.
03

Compute the number of photons per second

Lastly, to calculate the number of photons per second, divide the power of the signal by the energy of a singular photon. Hence, using the signal power \(4 \times 10^{-21} W\) (where 1W = 1 joule/sec), the number of photons per second can be calculated as: \(4 \times 10^{-21}\) divided by \(5.57 \times 10^{-24}\) = \(7.18 \times 10^{2}\) photons per second.

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