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One form of X-band radar has a frequency of 8.0 GHz , which is the same as \( 8.0 \times 10^{9} \mathrm{~Hz} \). What is the energy of the radar? \( \mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s} \) \( [?] \times 10^{[?]} \mathrm{J} \)

Ask by Schofield Chadwick. in the United States
Jan 24,2025

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Tutor-Verified Answer

Answer

The energy of the radar is \( 5.3 \times 10^{-24} \, \mathrm{J} \).

Solution

To calculate the energy \( E \) of a single photon of the radar wave, we can use the formula: \[ E = h \nu \] where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \mathrm{J \cdot s} \)) - \( \nu \) (nu) is the frequency of the radar (\( 8.0 \times 10^{9} \, \mathrm{Hz} \)) Plugging in the values: \[ E = 6.626 \times 10^{-34} \, \mathrm{J \cdot s} \times 8.0 \times 10^{9} \, \mathrm{Hz} \] \[ E = 5.3008 \times 10^{-24} \, \mathrm{J} \] Rounding to two significant figures: \[ E \approx 5.3 \times 10^{-24} \, \mathrm{J} \] **Answer:** \( 5.3 \times 10^{-24} \mathrm{~J} \)

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To find the energy of the radar, we can use the formula for energy related to frequency: \[ E = h \cdot f \] where \(E\) is the energy in joules, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}\)), and \(f\) is the frequency in hertz (\(8.0 \times 10^{9} \mathrm{~Hz}\)). Plugging in the values: \[ E = 6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s} \cdot 8.0 \times 10^{9} \mathrm{~Hz} \] Calculating this gives: \[ E = 5.3008 \times 10^{-24} \mathrm{~J} \] So, the energy of the radar is approximately \(5.3 \times 10^{-24} \mathrm{~J}\). The answer is: \(5.3 \times 10^{-24} \mathrm{~J}\)

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