Problem 4 - College Physics

Question

A radio station broadcasts at a frequency of \(92.0 \mathrm{MHz}\) with a power output of \(50.0 \mathrm{~kW}\). (a) What is the energy of each emitted photon, in joules and in electronvolts? (b) How many photons are emitted per second?

Step-by-Step Solution

Answer

(a) Each photon has an energy of \(6.10 \times 10^{-26}\, \text{J}\) or \(3.81 \times 10^{-7}\, \text{eV}\). (b) Approximately \(8.20 \times 10^{29}\) photons are emitted per second.

Step 1: Convert frequency from MHz to Hz

The frequency given is in megahertz (MHz), which needs to be converted to hertz (Hz) to use in calculations. Since 1 MHz is equal to \( 10^6 \) Hz, we have:\[92.0 \, \text{MHz} = 92.0 \times 10^6 \, \text{Hz} = 9.2 \times 10^7 \, \text{Hz}\]

Step 2: Calculate energy of a photon in joules

The energy of a photon is calculated using the equation \( E = h \cdot f \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \) and \( f \) is the frequency:\[E = 6.626 \times 10^{-34} \, \text{Js} \times 9.2 \times 10^7 \, \text{Hz} = 6.09992 \times 10^{-26} \, \text{J}\]

Step 3: Convert energy from joules to electronvolts

To convert energy from joules to electronvolts, use the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \):\[E = \frac{6.09992 \times 10^{-26} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 3.81 \times 10^{-7} \, \text{eV}\]

Step 4: Calculate number of photons emitted per second

The number of photons emitted per second can be determined by dividing the total power by the energy of a single photon. Given the power output \( P = 50.0 \, \text{kW} = 50.0 \times 10^3 \, \text{W} = 50.0 \times 10^3 \, \text{J/s} \):\[\text{Number of photons} = \frac{50.0 \times 10^3 \, \text{J/s}}{6.09992 \times 10^{-26} \, \text{J/photon}} \approx 8.20 \times 10^{29} \, \text{photons/s}\]

Key Concepts

Radio Frequency

Radio frequency is a crucial concept in the realm of electromagnetic waves. It refers to the oscillations per second of radio waves, typically measured in hertz (Hz). Since radio waves are part of the electromagnetic spectrum, their frequency can vary widely, from kilohertz (kHz) for AM radio to gigahertz (GHz) for satellite communications.
In this exercise, the frequency was initially given in megahertz (MHz), which is a common unit for radio stations. To perform scientific calculations, we converted it to hertz by multiplying by 10⁶. This conversion is vital because calculations involving frequency, like calculating photon energy, require consistent units.

Planck's Constant

Planck’s constant is a fundamental constant in physics represented by the symbol \( h \). It has a value of approximately \( 6.626 \, \times \, 10^{-34} \, \text{Js} \). This small number plays a massive role in quantum mechanics and electromagnetic theory.
When determining the energy of a photon, Planck's constant is used in the simple formula \( E = h \, \cdot \, f \), where \( f \) is the frequency. This formula highlights the relationship between the energy of a photon and its frequency. In the exercise, Planck's constant helps us calculate the energy carried by each photon broadcasted by the radio station at a specific frequency.

Power Output

Power output, measured in watts (W), represents the rate at which energy is emitted, consumed, or converted by a source. A radio station's power output tells us how much energy it discharges every second. In the context of the exercise, a power output of 50.0 kW (kilowatts) translates to 50,000 joules per second.
This power output influences how many photons are emitted by the radio station. By understanding this relationship, we can calculate the number of photons by dividing the total energy output (in joules per second) by the energy of a single photon (also in joules). Such insights are essential for analyzing how different power levels affect transmission strength and clarity.

Frequency Conversion

Frequency conversion is an essential step in many physics calculations, particularly when dealing with electromagnetic waves. Since frequency can be expressed in various units like hertz, kilohertz, megahertz, and gigahertz, converting these to a standard unit (usually hertz) ensures accuracy in calculations.
In the given exercise, converting from MHz to Hz was crucial for calculating the energy of a photon correctly. By knowing that 1 MHz equals \( 10^6 \) Hz, we aligned the frequency units with Planck's constant for precise calculations. Without this step, our photon energy calculations using \( E = h \, \cdot \, f \) would not yield accurate results. Thus, frequency conversion underpins accurate scientific problem-solving involving waves.