Ch 38: Quantization

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 38: Quantization

Problem 44

Chapter 38, Problem 44

A 75 kW radio transmitter emits 550 kHz radio waves uniformly in all directions. At what rate do photons strike a 1.5-m-tall, 3.0-mm-diameter antenna that is 15 km away?

Verified step by step guidance

Step 1: Calculate the energy of a single photon using the formula \( E = h \cdot f \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J·s}) \) and \( f \) is the frequency of the radio wave \( (550 \times 10^{3} \, \text{Hz}) \).

Step 2: Determine the total power emitted by the transmitter, which is given as \( P = 75 \, \text{kW} \) or \( 75 \times 10^{3} \, \text{W} \). This represents the energy emitted per second.

Step 3: Calculate the number of photons emitted per second using \( N = \frac{P}{E} \), where \( P \) is the total power and \( E \) is the energy of a single photon.

Step 4: Compute the intensity of the radio waves at a distance of \( r = 15 \, \text{km} \) using the formula \( I = \frac{P}{4 \pi r^2} \), where \( r \) is the distance from the transmitter. Convert \( r \) to meters before substituting.

Step 5: Calculate the rate at which photons strike the antenna by multiplying the intensity \( I \) by the cross-sectional area of the antenna \( A \), where \( A = \pi \cdot \left( \frac{d}{2} \right)^2 \) and \( d \) is the diameter of the antenna (3.0 mm converted to meters).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power and Energy of Photons

The power of a transmitter, measured in watts, indicates the rate at which it emits energy. Each photon carries energy determined by its frequency, given by the equation E = hf, where E is energy, h is Planck's constant, and f is frequency. Understanding the relationship between power and photon energy is essential for calculating the number of photons emitted per second.

Intensity of Electromagnetic Waves

The intensity of an electromagnetic wave is defined as the power per unit area. It decreases with distance from the source according to the inverse square law, which states that intensity is proportional to 1/r², where r is the distance from the source. This concept is crucial for determining how many photons reach the antenna based on its distance from the transmitter.

Photon Flux

Photon flux refers to the number of photons passing through a given area per unit time. It can be calculated by dividing the intensity of the electromagnetic wave by the energy of a single photon. This concept is vital for finding the rate at which photons strike the antenna, as it directly relates the intensity of the radio waves to the physical dimensions of the antenna.

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Textbook Question

INT The electron interference pattern of Figure 38.12 was made by shooting electrons with 50 keV of kinetic energy through two slits spaced 1.0 μm apart. The fringes were recorded on a detector 1.0 m behind the slits. Figure 38.12 is greatly magnified. What was the actual spacing on the detector between adjacent bright fringes?

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Textbook Question

Potassium and gold cathodes are used in a photoelectric-effect experiment. For each cathode, find: The threshold frequency.

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Textbook Question

An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 eV, 27 eV, and 48 eV. What is the length of the box?

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Textbook Question

The graph in FIGURE P38.42 was measured in a photoelectric-effect experiment. What is the work function (in eV) of the cathode?

Textbook Question

Potassium and gold cathodes are used in a photoelectric-effect experiment. For each cathode, find: The stopping potential if the wavelength is 220 nm.

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Textbook Question

The cosmic microwave background radiation is light left over from the Big Bang that has been Doppler-shifted to microwave frequencies by the expansion of the universe. It now fills the universe with 450 photons/cm³ at an average frequency of 160 GHz. How much energy from the cosmic microwave background, in MeV, fills a small apartment that has 95 m² of floor space and 2.5-m-high ceilings?

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