To find the number of photons emitted per second by a radio transmitter operating at a frequency of 880 kHz and a power of 10 kW, we can follow these steps:
### Step 1: Understand the relationship between power, energy, and number of photons
Power (P) is defined as the rate of energy transfer. It can be expressed as:
\[ P = \frac{E}{t} \]
where \( E \) is the energy transferred and \( t \) is the time.
The number of photons emitted per second (n) can be expressed in terms of power and the energy of a single photon:
\[ P = n \cdot E_{photon} \]
### Step 2: Calculate the energy of a single photon
The energy of a single photon can be calculated using the formula:
\[ E_{photon} = h \cdot \nu \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J s} \))
- \( \nu \) is the frequency of the radiation in Hertz (Hz).
Given that the frequency \( \nu \) is \( 880 \, \text{kHz} \), we convert it to Hertz:
\[ \nu = 880 \, \text{kHz} = 880 \times 10^3 \, \text{Hz} \]
Now, we can calculate the energy of a single photon:
\[ E_{photon} = 6.63 \times 10^{-34} \, \text{J s} \times 880 \times 10^3 \, \text{Hz} \]
### Step 3: Calculate the energy of a single photon
Calculating this gives:
\[ E_{photon} = 6.63 \times 10^{-34} \times 880 \times 10^3 \]
\[ E_{photon} = 5.83 \times 10^{-28} \, \text{J} \]
### Step 4: Convert power to watts
The power of the transmitter is given as \( 10 \, \text{kW} \). We convert this to watts:
\[ P = 10 \, \text{kW} = 10 \times 10^3 \, \text{W} = 10^4 \, \text{W} \]
### Step 5: Calculate the number of photons emitted per second
Using the relationship \( P = n \cdot E_{photon} \), we can rearrange to find \( n \):
\[ n = \frac{P}{E_{photon}} \]
Substituting the values we have:
\[ n = \frac{10^4 \, \text{W}}{5.83 \times 10^{-28} \, \text{J}} \]
### Step 6: Perform the calculation
Calculating this gives:
\[ n \approx 1.72 \times 10^{31} \]
### Final Answer
The number of photons emitted per second is approximately:
\[ n \approx 1.72 \times 10^{31} \]
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