Question

A radio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is

• A. $1.72 \times 10^{31}$
• B. $1.327 \times 10^{25}$
• C. $1.327 \times 10^{37}$
• D. $1.327 \times 10^{45}$

Answer 1

The Correct Option is A

To determine the number of photons emitted per second by the radio transmitter, we need to use the relationship between energy, frequency, and the number of photons. The relevant formula is:

E = n \cdot h \cdot f

Here, E represents the total energy emitted per second (the power of the transmitter), n is the number of photons, h is Planck's constant (6.626 \times 10^{-34} \text{ J s}), and f is the frequency of the radio waves.

1. Identify the Given Values:
   • Frequency (f): 880 \text{ kHz} = 880 \times 10^3 \text{ Hz}
   • Power (E): 10 \text{ kW} = 10 \times 10^3 \text{ W}
2. Use the Formula to Calculate Number of Photons:

   Rearrange the formula E = n \cdot h \cdot f to solve for n:

   n = \frac{E}{h \cdot f}

3. Substitute the Values:

   Substitute the given values into the formula:

   n = \frac{10 \times 10^3}{6.626 \times 10^{-34} \times 880 \times 10^3}

   Calculate the result:

   n = \frac{10 \times 10^3}{6.626 \times 880} \times 10^{-31}

   n \approx \frac{10 \times 10^3}{5820.88} \times 10^{-31} \approx 1.72 \times 10^{31}

Therefore, the number of photons emitted per second by the radio transmitter is approximately 1.72 \times 10^{31}. The correct option is the first one.