Question

A 1000 watt radio transmitter operates at a frequen cy of \(880 \mathrm{kc} / \mathrm{sec}\). How many photons per sec. does it emit? \(\left[\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right]\) (a) \(2.51 \times 10^{30}\) (b) \(2.27 \times 10^{28}\) (c) \(1.72 \times 10^{30}\) (d) \(1.77 \times 10^{27}\)

Short answer

The number of photons emitted per second is approximately \(1.77 \times 10^{27}\), option (d).

Step-by-step solution

Understanding the Problem

We need to find the number of photons emitted by a 1000-watt radio transmitter operating at a frequency of \(880 \text{kc/sec}\) (where \(\text{k} = 10^3\)). This involves determining how much energy each photon carries and dividing the total energy output per second (1000 watts) by the energy per photon.

Convert Frequency to Hertz

The given frequency is \(880 \text{kc/sec}\), which stands for kilocycles per second or kilohertz. We convert this frequency into hertz (Hz): \(880 \text{ kHz} = 880 \times 10^3 \text{ Hz}\).

Calculate Energy per Photon

Use Planck's equation \(E = h \cdot u\) to find the energy of a single photon. Here, \(h = 6.626 \times 10^{-34} \text{ J s}\) and \(u = 880 \times 10^3 \text{ Hz}\). So, the energy per photon is \(E = 6.626 \times 10^{-34} \times 880 \times 10^3\).

Calculate Total Number of Photons Emitted

The transmitter emits 1000 joules per second (since 1000 watts = 1000 joules/sec). To find out how many photons this corresponds to, divide the total energy output per second by the energy per photon. This gives us the number of photons emitted per second: \(\text{Number of photons} = \frac{1000}{6.626 \times 10^{-34} \cdot 880 \times 10^3}\).

Simplify and Compute

Calculate the expression from Step 4: \(\text{Number of photons} = \frac{1000}{6.626 \times 10^{-34} \times 8.8 \times 10^5}\). This simplifies to approximately \(1.77 \times 10^{27}\).

Key concepts

Planck's Equation

The exploration of photon emission starts with understanding Planck's Equation. This equation is a fundamental element in quantum mechanics and describes the quantization of energy levels. It is expressed as: \[ E = h \cdot u \] where

• \(E\) is the energy of a photon.
• \(h\) is Planck's constant, valued at 6.626 x 10^{-34} J·s.
• \(u\) is the frequency of the electromagnetic radiation in hertz (Hz).

Planck’s equation, therefore, establishes that the energy carried by a photon is directly proportional to the frequency of the radiation. The higher the frequency, the greater the energy per photon. This relationship is incredibly significant when determining parameters like energy emission rates or power distributions in various applications, from radio transmitters to spectral analysis in scientific research.

Frequency Conversion

In the context of electromagnetic waves, frequency conversion is crucial as it allows us to use correct units for solving physics problems. Original frequency may be listed in various forms, such as kilohertz (kHz), megahertz (MHz), or gigahertz (GHz). Converting these to hertz (Hz) is necessary because standard equations, like Planck's, operate using hertz. For instance, the given problem provides frequency in kilocycles per second (\(kc/sec\)), or kilohertz. To convert 880 \(kc/sec\) to hertz:

• Understand that 1 kHz = 1,000 Hz.
• Multiply 880 by 1,000, resulting in 880,000 Hz or \(8.8 \times 10^5\) Hz.

This conversion ensures that units are consistent with those in Planck's Equation, thereby enabling an accurate calculation of photon energy. Remember, unit consistency is a fundamental rule in physics that ensures dimensional insight and accurate results.

Energy per Photon Calculation

After establishing the basics of Planck's equation and converting frequency to the appropriate units, the next logical step is calculating the energy each photon carries. This is essential for assessing how many photons are generated by a source of given energy output, like the 1000-watt radio transmitter from the problem. Let’s revisit the Planck’s equation: \[ E = 6.626 \times 10^{-34} \cdot 8.8 \times 10^5 \] Calculating the result of this equation gives:

• The energy of a single photon is approximately \(5.8328 \times 10^{-28}\) joules.

With this energy per photon, you can determine how many photons correspond to the radio transmitter's total energy output. Divide the transmitter's total energy output per second (1000 joules) by the energy per individual photon. This provides the photon emission rate per second: \[ \text{Number of photons} = \frac{1000}{5.8328 \times 10^{-28}} \approx 1.77 \times 10^{27} \]Through this simple yet profound calculation, the concept of photon-based energy quantification becomes intuitively clear, allowing for enhanced understanding of emissions from light sources.