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The radio transmitter operates at an frequency of \(880 \mathrm{kHz}\) and a power of \(10 \mathrm{~kW}\). The number of photons per second emitted are (a) \(1.171 \times 10^{25}\) (b) \(1.715 \times 10^{10}\) (c) \(1.71 \times 10^{91}\) (d) \(1.025 \times 10^{\% 0}\)

Short Answer

Expert verified
None of the given options match; the number is approximately \(1.715 \times 10^{31}\).

Step by step solution

01

Determine the Energy of a Single Photon

To find the energy of a single photon, use the formula: \[ E = h \cdot f \] where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J} \cdot \text{s}) \) and \( f \) is the frequency of the radio wave. Given \( f = 880 \text{kHz} = 880 \times 10^3 \text{ Hz}, \) we calculate:\[ E = 6.626 \times 10^{-34} \times 880 \times 10^3 \] \[ E = 5.833 \times 10^{-28} \text{ J} \] Thus, the energy of a single photon is \( 5.833 \times 10^{-28} \text{ Joules}. \)
02

Calculate Total Power in Joules Per Second

The power given is in kilowatts, which needs to be converted to watts (joules per second) to match the energy units. \[ P = 10 \text{ kW} = 10 \times 10^3 \text{ W} = 10^4 \text{ J/s} \] Hence, the transmitter operates at a power of \( 10^4 \text{ Joules per second}.\)
03

Calculate Number of Photons Emitted Per Second

To find the number of photons emitted per second, divide the total power by the energy of one photon:\[ n = \frac{P}{E} = \frac{10^4}{5.833 \times 10^{-28}} \] \[ n \approx 1.715 \times 10^{31} \] This number represents the number of photons emitted by the transmitter per second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's constant
Planck's constant, denoted as \( h \), is a fundamental constant in physics that plays a crucial role in the quantum mechanical description of the energy and momentum of particles. Its value is \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \). This constant represents the smallest possible action or energy unit in the quantum realm. Discovered by Max Planck, it serves as a bridge between the wave and particle natures of light. It is vital in calculations involving the energy of photons, which are small packets of light energy. The relationship between energy and frequency of a photon is represented by the equation \( E = h \cdot f \), where \( E \) is the energy and \( f \) is the frequency. This shows how quantized energy levels emerge—energy can't be continuous but comes in discrete chunks defined by \( h \). Understanding this principle is essential for exploring quantum mechanics and the nature of electromagnetic radiation.
Energy calculation
The calculation of energy of a single photon involves multiplying Planck's constant \( h \) by the frequency \( f \) of the photon using the equation \( E = h \cdot f \). It is important to know the correct units:
  • Planck's constant \( h \) is in Joule seconds (J \( \cdot \) s)
  • Frequency \( f \) is in hertz (Hz), which is equivalent to cycles per second

When calculating the energy of a photon associated with a given frequency, first make sure to convert all units appropriately. For instance, in the exercise, the frequency is given in kilohertz (kHz), which must be converted to hertz by multiplying by \( 10^3 \). As demonstrated, \( 880 \text{kHz} = 880 \times 10^3 \text{Hz} \). Calculating the energy using this frequency yields \( E = 5.833 \times 10^{-28} \text{J} \). Recognizing such minute amounts of energy underscores the fact that photons, while impressively powerful in aggregate, carry very small energy individually.
Frequency and wavelength
Frequency and wavelength are fundamental concepts that describe the properties of waves, including light waves and sound waves. The frequency \( f \) of a wave is the number of oscillations or cycles that occur in a second, measured in hertz (Hz). Conversely, wavelength \( \lambda \) is the distance between consecutive crests (or troughs) of a wave. The relationship between frequency and wavelength is inversely proportional and is governed by the equation \( c = \lambda \cdot f \), where \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \text{ m/s} \). As the frequency of a wave increases, its wavelength decreases, and vice versa.
  • Higher frequency = shorter wavelength
  • Lower frequency = longer wavelength

This interplay between frequency and wavelength is a core aspect of electromagnetic theory, allowing us to determine properties of waves when only given frequency or wavelength. In practical scenarios, understanding these concepts allows electronics, like radio transmitters, to operate effectively at specified frequencies and wavelengths.

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Most popular questions from this chapter

The sum radiates electromagnetic energy at the rate of \(4.0 \times 10^{26} \mathrm{~W}\). Its radius is \(7 \times 10^{8} \mathrm{~m}\). The intensity of sun light at the solar surface is (a) \(4.54 \times 10^{8} \mathrm{Wm}^{-2}\) (b) \(5.09 \times 10^{-8} \mathrm{Wm}^{-2}\) (c) \(3.00 \times 10^{8} \mathrm{Wm}^{-2}\) (d) \(10^{\mathrm{E}} \mathrm{Wm}^{-1}\)

A radiation of \(200 \mathrm{~W}\) is incident on a surface which is \(60 \%\) reflecting and \(40 \%\) absorbing. The total force on the surface is (a) \(107 \times 10^{-6} \mathrm{~N}\) (b) \(13 \times 10^{-6} \mathrm{~N}\) (c) \(107 \times 10^{-7} \mathrm{~N}\) (d) \(103 \times 10^{-7} \mathrm{~N}\)

If the wavelength of light is \(4000 \AA\), then the number of wavelength in \(1 \mathrm{~mm}\) length will be (a) \(2.5 \times 10^{5}\) (b) \(0.25 \times 10^{4}\) (c) \(2.5 \times 10^{4}\) (d) 0

A plane electromagnetic wave of intensity \(10 \mathrm{Wm}^{-2}\) strikes a small mirror of area \(20 \mathrm{~cm}^{2}\), held perpendicular to the approaching wave. The radiation force on the mirror will (a) \(6.6 \times 10^{-11} \mathrm{~N}\) (b) \(133 \times 10^{-11} \mathrm{~N}\) (c) \(133 \times 10^{-10} \mathrm{~N}\) (d) \(6.6 \times 10^{-11} \mathrm{~N}\)

A plane electromagnetic waves travelling along the \(x\)-direction has a wavelength of \(3 \mathrm{~mm}\). The variation in the electric field occurs in the \(y\)-direction with an amplitude \(66 \mathrm{Vm}^{-1}\). The equation for the electric and magnetic fields as a function of \(x\) and \(t\) are respectively (a) \(E_{y}=33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{2}=1.1 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) (b) \(E_{y}=11 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{y}=11 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) (c) \(E_{y}=33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{y}=11 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) (d) \(E_{y}=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{z}=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\)

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