A 1000 W transmitter works at a frequency of 880kHz. The number of photons emitted per second Is

Options

A. `1.7xx10^28`
B. `1.7xx10^30`
C. `1.7xx10^23`
D. `1xx10^25`

Explanation

To find the number of photons emitted per second by a 1000 W transmitter operating at a frequency of 880 kHz, we can use the formula:

\[ N = \frac{P}{h \cdot \nu} \]

where:
- \( N \) is the number of photons emitted per second,
- \( P \) is the power of the transmitter (in watts),
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( \nu \) is the frequency (in hertz).

### Step 1: Convert the frequency from kHz to Hz
Given that the frequency is 880 kHz, we convert it to Hz:

\[ \nu = 880 \, \text{kHz} = 880 \times 10^3 \, \text{Hz} = 8.8 \times 10^5 \, \text{Hz} \]

### Step 2: Substitute the values into the formula
Now we can substitute the values into the formula. The power \( P \) is 1000 W, \( h \) is \( 6.626 \times 10^{-34} \, \text{J s} \), and \( \nu \) is \( 8.8 \times 10^5 \, \text{Hz} \):

\[ N = \frac{1000}{6.626 \times 10^{-34} \cdot 8.8 \times 10^5} \]

### Step 3: Calculate the denominator
First, we calculate the denominator:

\[ h \cdot \nu = 6.626 \times 10^{-34} \cdot 8.8 \times 10^5 \]

Calculating this gives:

\[ h \cdot \nu \approx 5.83 \times 10^{-28} \, \text{J} \]

### Step 4: Calculate the number of photons emitted per second
Now, substituting back into the equation for \( N \):

\[ N = \frac{1000}{5.83 \times 10^{-28}} \approx 1.72 \times 10^{30} \]

### Conclusion
Thus, the number of photons emitted per second by the transmitter is approximately:

\[ N \approx 1.72 \times 10^{30} \, \text{photons/s} \]

### Final Answer
The number of photons emitted per second is \( 1.72 \times 10^{30} \).

---