
An AIR station is broadcasting waves of wavelength 300 metres. If the radiating power of the transmitter is 10kW, then the number of photons radiated per second is:
A. \[1.5 \times {10^{29}}\]
B. \[1.5 \times {10^{31}}\]
C. \[1.5 \times {10^{33}}\]
D. \[1.5 \times {10^{35}}\]
Answer
219k+ views
Hint: Power of the wave is the energy per unit of time. When we find the total energy of the total photon per unit of time then we get the power of the wave. The photon is the energy packet of the wave.
Formula used:
\[E = \dfrac{{hc}}{\lambda }\]
where E is the energy per photon of wavelength \[\lambda \], h is the Plank’s constant and \[c\] is the speed of light.
\[{E_t} = nE\]
where E is the energy per photon, \[{E_t}\] is the total energy of the wave which contains a total of n photons.
Complete step by step solution:
The wavelength of the wave is given as 300 metres. Using the given wavelength, we find the energy which each photon carries,
\[E = \dfrac{{hc}}{\lambda }\]
\[\Rightarrow E = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{300}}J\]
\[\Rightarrow E = 6.626 \times {10^{ - 28}}J\]
Hence, each photon contains \[6.626 \times {10^{ - 28}}J\] energy.
If there are n photons per second, then total energy of the wave per second will be equal to the power of the wave, i.e. \[P = nE\]. So, the total energy of the wave per second is,
\[{E_t} = nE\]
The expression for the number of the photons will be,
\[n = \dfrac{{{E_t}}}{E}\]
As the total energy of the wave per second is equal to the power of the wave, so \[{E_t} = 10kJ\] per second.
\[{E_t} = 10 \times {10^3}J\]
Putting the values in the expression for the number of photons per unit second, we get
\[n = \dfrac{{10 \times {{10}^3}}}{{6.626 \times {{10}^{ - 28}}}}\]
\[\therefore n = 1.5 \times {10^{31}}\]
Hence, the wave with a wavelength of 300 metres and having the power of 10 kW contains \[1.5 \times {10^{31}}\] photons per second.
Therefore, the correct option is B.
Note: The rest mass of a photon is zero. The photon is a quantization of the energy of an electromagnetic wave. The momentum of a particle is the characteristic of motion. When an electromagnetic wave travels then it carries energy in the form of energy packets called photons.
Formula used:
\[E = \dfrac{{hc}}{\lambda }\]
where E is the energy per photon of wavelength \[\lambda \], h is the Plank’s constant and \[c\] is the speed of light.
\[{E_t} = nE\]
where E is the energy per photon, \[{E_t}\] is the total energy of the wave which contains a total of n photons.
Complete step by step solution:
The wavelength of the wave is given as 300 metres. Using the given wavelength, we find the energy which each photon carries,
\[E = \dfrac{{hc}}{\lambda }\]
\[\Rightarrow E = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{300}}J\]
\[\Rightarrow E = 6.626 \times {10^{ - 28}}J\]
Hence, each photon contains \[6.626 \times {10^{ - 28}}J\] energy.
If there are n photons per second, then total energy of the wave per second will be equal to the power of the wave, i.e. \[P = nE\]. So, the total energy of the wave per second is,
\[{E_t} = nE\]
The expression for the number of the photons will be,
\[n = \dfrac{{{E_t}}}{E}\]
As the total energy of the wave per second is equal to the power of the wave, so \[{E_t} = 10kJ\] per second.
\[{E_t} = 10 \times {10^3}J\]
Putting the values in the expression for the number of photons per unit second, we get
\[n = \dfrac{{10 \times {{10}^3}}}{{6.626 \times {{10}^{ - 28}}}}\]
\[\therefore n = 1.5 \times {10^{31}}\]
Hence, the wave with a wavelength of 300 metres and having the power of 10 kW contains \[1.5 \times {10^{31}}\] photons per second.
Therefore, the correct option is B.
Note: The rest mass of a photon is zero. The photon is a quantization of the energy of an electromagnetic wave. The momentum of a particle is the characteristic of motion. When an electromagnetic wave travels then it carries energy in the form of energy packets called photons.
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