Question

An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?

Answer 1

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = $\frac{NE}{t}$

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = $\frac{NE}{t}$ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$          [t = 1 s for per second]

or

550 × 10³ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

Answer 2

The number of photons that are emitted by this AM radio transmitter is equal to $1.12 \times 10^{33}\;photons.$

Given the following data:

• Power = 550 kW.

• Frequency = 740 kHz.

Scientific data:

• Planck constant = $6.626 \times 10^{-34}\;J.s$

How to calculate the number of photons.

In order to determine the number of photons that are being emitted by this AM radio transmitter, we would solve for the quantity of energy it consumes by using Planck-Einstein's equation.

Mathematically, the Planck-Einstein relation is given by the formula:

$E = hf$

Where:

• h is Planck constant.

• f is photon frequency.

Substituting the given parameters into the formula, we have;

$E = 6.626 \times 10^{-34}\times 740 \times 10^3\\\\E = 4.903 \times 10^{-28}\;Joules.$

For the number of photons:

$n=\frac{Power}{Energy} \\\\n=\frac{550 \times 10^3}{4.903 \times 10^{-28}} \\\\n=1.12 \times 10^{33}\;photons.$

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