Question

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

Answer 1

Step-1: Given data

1. The frequency of the radio transmitter is $f=880KHz=880\times {10}^{3}Hz.$
2. The power of the transmitter is $P=10kw=10\times {10}^{3}watt.$

Step-2: The formula to be used

1. Power is the amount of energy consumed by a body in a unit time. Power is defined by the form, $P=\frac{E}{t}$, where, E is the energy and t is the time.
2. The energy of a single photon is $E=hf$, where, h $\left(h=6.626\times {10}^{-34}j.s\right)$ is the Plank constant and f is the frequency of the photon.

Step-3: Finding the number of photons

As we know, power, $P=\frac{E}{t}$.

So, at unit time $P=E$.

Now, the number of photons per second, $n=\frac{Totalpowerofthetransmitter}{Powerof\sin glephoton}$

So,

$$\begin{gathered} n=\frac{Totalpowerofthetransmitter}{Powerof\sin glephoton}=\frac{P}{hf}=\frac{10\times {10}^3}{\left(6.626\times {10}^{-34}\right)\times 880\times {10}^3} \\ orn=1.7\times {10}^{31} \end{gathered}$$

Therefore, the number of photons emitted per second is $n=1.7\times {10}^{31}$