
A radio station emits \[10kW\]power of $90.8MHz$. Find the number of photons emitted per second.
A. $1.6 \times {10^{28}}$
B. $1.6 \times {10^{29}}$
C. $1.6 \times {10^{30}}$
D. $1.6 \times {10^{32}}$
Answer
544.8k+ views
Hint:To solve this question, we will use the concept energy of the photon. It is related to the planck's constant and frequency. After that we will use the relation between the number of photons, power and energy of the photon which will lead us to the final answer.
Formula used:
$E = hv$,
where, $E$ is the energy of the photon,$h$ is the planck's constant and $v$ is the frequency
$N = \dfrac{P}{E}$
where, $N$ is the number of photons emitted per second,$P$ is the power and $E$ is the energy of photon
Complete step by step answer:
We are given the power and the frequency of the radio station.
We know that the number of photons emitted per second is given by the formula
$N = \dfrac{P}{E}$
Also the energy of the photon is given by $E = hv$
Therefore, the number of photons emitted per second will be
$N = \dfrac{P}{{hv}}$
We are given that
$P = 10kW = 10 \times {10^3}W$
$\Rightarrow v = 90.8MHz = 90.8 \times {10^6}Hz$
Also the value of planck constant $h = 6.626 \times {10^{ - 34}}{m^2}kg/s$
Putting these values in the equation, we get
$\therefore N = \dfrac{{10 \times {{10}^3}}}{{6.626 \times {{10}^{ - 34}} \times 90.8 \times {{10}^6}}} = 1.6 \times {10^{29}}$
Thus, the number of photons emitted per second is $1.6 \times {10^{29}}$.
Hence, option B is the right answer.
Note:We have used the concept of energy of photon which is based on the photoelectric effect. The photoelectric effect cannot be explained by considering light as a wave. However, it can be explained by the particle nature of light, in which light can be visualized as a stream of particles of electromagnetic energy. These particles of light are called photons. The energy held by a photon is related to the frequency of the light via Planck’s equation $E = hv$.
Formula used:
$E = hv$,
where, $E$ is the energy of the photon,$h$ is the planck's constant and $v$ is the frequency
$N = \dfrac{P}{E}$
where, $N$ is the number of photons emitted per second,$P$ is the power and $E$ is the energy of photon
Complete step by step answer:
We are given the power and the frequency of the radio station.
We know that the number of photons emitted per second is given by the formula
$N = \dfrac{P}{E}$
Also the energy of the photon is given by $E = hv$
Therefore, the number of photons emitted per second will be
$N = \dfrac{P}{{hv}}$
We are given that
$P = 10kW = 10 \times {10^3}W$
$\Rightarrow v = 90.8MHz = 90.8 \times {10^6}Hz$
Also the value of planck constant $h = 6.626 \times {10^{ - 34}}{m^2}kg/s$
Putting these values in the equation, we get
$\therefore N = \dfrac{{10 \times {{10}^3}}}{{6.626 \times {{10}^{ - 34}} \times 90.8 \times {{10}^6}}} = 1.6 \times {10^{29}}$
Thus, the number of photons emitted per second is $1.6 \times {10^{29}}$.
Hence, option B is the right answer.
Note:We have used the concept of energy of photon which is based on the photoelectric effect. The photoelectric effect cannot be explained by considering light as a wave. However, it can be explained by the particle nature of light, in which light can be visualized as a stream of particles of electromagnetic energy. These particles of light are called photons. The energy held by a photon is related to the frequency of the light via Planck’s equation $E = hv$.
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