Question

1 kW radio transmitter operates at a frequency of 800 Hz. How many photons per second does it emit?
A. $1.89\times {{10}^{21}}$
B. $1.89\times {{10}^{33}}$
C. $6.02\times {{10}^{23}}$
D. $2.85\times {{10}^{20}}$

Answer 1

Hint: We know that energy is equal to the number of photons times Planck's constant times the frequency. If we divide the energy by Planck's constant, we get photons per emitted per second.

Complete Step by step solution:
Let us calculate the energy of one photon which has a frequency of 800 Hz.
We know that the energy of radiation or photon is given by: \[E=h\upsilon \], where h is the Planck's constant and \[\upsilon \] is the frequency of the photon.
Planck's constant =\[6.625\times {{10}^{-34}}Js\] and frequency of photon = of 800 Hz
Substituting in the energy equation we get:
\[\begin{array}{*{35}{l}}
   E\text{ }\!\!~\!\!\text{ }=\text{ }\!\!~\!\!\text{ }6.625\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }{{10}^{-34}}\text{ }\!\!~\!\!\text{ }J-sec\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }800\text{ }\!\!~\!\!\text{ }se{{c}^{-1}} \\
   E\text{ }\!\!~\!\!\text{ }=\text{ }\!\!~\!\!\text{ }5.3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }{{10}^{-31}}\text{ }\!\!~\!\!\text{ }Joules. \\
\end{array}\]
The energy of one photon which has frequency 800 Hz is \[E\text{ }\!\!~\!\!\text{ }=\text{ }\!\!~\!\!\text{ }5.3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }{{10}^{-31}}\text{ }\!\!~\!\!\text{ }Joules\].
Given the power of the radio transmitter as 1 kW = 1000 watts = 1000 \[J/sec\].
(As we know that 1 watt = 1 joule per second)
The radio transmitter emits 1000 Joules in one second, but we know that the energy of one photon is \[5.3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }{{10}^{-31}}\text{ }\!\!~\!\!\text{ }Joules\].
If the energy of one photon is \[5.3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }{{10}^{-31}}\text{ }\!\!~\!\!\text{ }Joules\] then 1000 joules contains how many photons are what we need to calculate.
A number of photons which make up the energy of 1000 joules are: \[\dfrac{\left( 1000\text{ }\!\!~\!\!\text{ }joules\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }1\text{ }\!\!~\!\!\text{ }photon \right)}{\left( 5.3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }{{10}^{-31}}\text{ }\!\!~\!\!\text{ }Joules \right)}\]
=\[1.89\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }10{}^\text{3}{}^\text{3}\text{ }\!\!~\!\!\text{ }photons\]
The radio transmitter of 1kW power which operates at a frequency of 800 Hz emits\[1.89\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }10{}^\text{3}{}^\text{3}\text{ }\!\!~\!\!\text{ }photons\] per second.

According to the above calculations, we conclude that an accurate answer to the given question is an option (B) \[1.89\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }10{}^\text{3}{}^\text{3}\].

Note: The SI unit for frequency is the hertz (Hz), one hertz is the same as one cycle per second and the SI unit of power is watt which is equal to one joule per second. Other common and traditional measures are horsepower (hp), one mechanical horsepower equals about 745.7 watts.