Question

the no.of photons emitted per second by a medium wave transmitter of 10kW power,emitting radio waves of wave length 500m

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Answer 1

Explanation:

Power=n*energy /time

E=h*frequency sub

n=power*wave length /h*c

n=10*500/6.6*10power-34*3*10power8n=0.25*10power29

Answer 2

Answer:

No.of photons emitted per second =$2.5153\times10^{31}$

Explanation:

Since we know that

$P=\frac{nE}{t}$.............................(1)

where P=power, n=number of photns, E=energy, t=time

we have

$E=\frac{hc}{Wavelength}$

h=planks constant=$6.626\times10^{-34}$,

c=speed of light=$3\times10^{8}$ and λ=wavelength.

we have wavelength of 500m. so energy is

$E=\frac{6.626\times10^{-34}\times3\times10^{8}  }{500}$

$E=\frac{1.9878\times10^{-25} }{500}$

$E=3.9756\times10^{-28}$

we have P=10000 watt.

so using (1) we get,

$10000=\frac{3.9756\times10^{-28} \times n}{1}$

$n=\frac{10000}{3.9756\times10^{-28} }$

So no.of photons emitted per second =$2.5153\times10^{31}$