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