Calculate the number of photons emitted by a radio tower of 150 kW that operates at a 99.7 MHz frequency.
Calculate the number of photons emitted by a radio tower of 150 kW that operates at a 99.7 MHz frequency.
Related questions
Question
100%
![### Problem Statement:
Calculate the number of photons emitted by a radio tower of 150 kW that operates at a 99.7 MHz frequency.
### Analysis:
In this problem, you are asked to find the number of photons emitted by a radio tower with a given power output and operating frequency.
### Solution:
First, recall the key relations involving photon energy and power:
1. **Energy of a single photon (E):**
\[ E = h \cdot f \]
where \( h \) is Planck's constant (approximately \( 6.626 \times 10^{-34} \) Js) and \( f \) is the frequency in Hz.
2. **Total power emitted (P):**
\[ P = N \cdot E \]
where \( N \) is the number of photons emitted per second.
Given:
- Power output (\( P \)) = 150 kW = \( 150 \times 10^3 \) W
- Frequency (\( f \)) = 99.7 MHz = \( 99.7 \times 10^6 \) Hz
##### Step-by-Step Calculation:
1. **Calculate the energy of a single photon:**
\[ E = h \cdot f \]
\[ E = (6.626 \times 10^{-34} \, \text{Js}) \cdot (99.7 \times 10^6 \, \text{Hz}) \]
\[ E \approx 6.603 \times 10^{-26} \, \text{J} \]
2. **Find the number of photons emitted per second:**
\[ P = N \cdot E \]
\[ N = \frac{P}{E} \]
\[ N = \frac{150 \times 10^3 \, \text{W}}{6.603 \times 10^{-26} \, \text{J}} \]
\[ N \approx 2.27 \times 10^{31} \]
### Conclusion:
The radio tower emits approximately \( 2.27 \times 10^{31} \) photons per second.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fe8677b-d2b4-4cf1-b1ed-08820154fcb5%2Fa05e411c-bff1-498e-8d52-e91fc837263c%2Fxrekkm8_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Calculate the number of photons emitted by a radio tower of 150 kW that operates at a 99.7 MHz frequency.
### Analysis:
In this problem, you are asked to find the number of photons emitted by a radio tower with a given power output and operating frequency.
### Solution:
First, recall the key relations involving photon energy and power:
1. **Energy of a single photon (E):**
\[ E = h \cdot f \]
where \( h \) is Planck's constant (approximately \( 6.626 \times 10^{-34} \) Js) and \( f \) is the frequency in Hz.
2. **Total power emitted (P):**
\[ P = N \cdot E \]
where \( N \) is the number of photons emitted per second.
Given:
- Power output (\( P \)) = 150 kW = \( 150 \times 10^3 \) W
- Frequency (\( f \)) = 99.7 MHz = \( 99.7 \times 10^6 \) Hz
##### Step-by-Step Calculation:
1. **Calculate the energy of a single photon:**
\[ E = h \cdot f \]
\[ E = (6.626 \times 10^{-34} \, \text{Js}) \cdot (99.7 \times 10^6 \, \text{Hz}) \]
\[ E \approx 6.603 \times 10^{-26} \, \text{J} \]
2. **Find the number of photons emitted per second:**
\[ P = N \cdot E \]
\[ N = \frac{P}{E} \]
\[ N = \frac{150 \times 10^3 \, \text{W}}{6.603 \times 10^{-26} \, \text{J}} \]
\[ N \approx 2.27 \times 10^{31} \]
### Conclusion:
The radio tower emits approximately \( 2.27 \times 10^{31} \) photons per second.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps
