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Radio transmitter radiates 1KW power at a wavelength 198.6m. How many photons does it emit per second?
A. ${10^{10}}$
B. ${10^{20}}$
C. ${10^{30}}$
D. ${10^{40}}$

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Last updated date: 25th Jul 2024
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Answer
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Hint:-
The best approach to solve the question is by using Einstein’s formula of energy for wave and particle nature that is,
$E = hv$, where h is the Planck’s constant whose value is fixed and equal to $6.626 \times {10^{ - 34}}JH{z^{ - 1}}$, E is the energy and $$v$$ is the frequency.
If consider $nE$ which means number of particles multiplied by energy which gives total energy and this total energy is used in the formula of Einstein.
We will also be going to use the relation of power and energy which is $P = \dfrac{E}{t}$, E is the energy and t is the time.

Complete step-by-step solution:Let us now calculate the Energy of one photon which has wavelength of 198.6m
As we know, $c = v\lambda $ where $v$ is the frequency in Hz, c is the speed of light which is $3 \times {10^8}$m/s and $\lambda $ is the wavelength in m
Putting the values in the formula,
$\eqalign{
  & 3 \times {10^8} = v \times 198.6 \cr
  & \Rightarrow v = 1.51 \times {10^6}Hz \cr} $
Now energy of one photon is,
$E = hv$
Putting the values of frequency and Planck’s constant,
$\eqalign{
  & E = 6.626 \times {10^{ - 34}} \times 1.51 \times {10^6} \cr
  & \Rightarrow E \approx 10 \times {10^{ - 28}}J \cr} $
Now power is total energy per unit time
Here time given is 1 second
So, $P = nE$ where n is the total number of particles
Putting the values given in question
$\eqalign{
  & 1000 = {10^{ - 28}} \times n \cr
  & \Rightarrow n = {10^{31}} \approx {10^{30}} \cr} $
Hence the correct option is (C).

Note:- Photon does not have mass so the way to calculate energy is different from other particles. The energy of a single photon is entirely dependent upon its frequency.
Make sure that you always use formulas with right dimensions and to remember the value of Planck’s constant though most of the time it is given but still learning will increase your practice.