Question

A $1$ 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

The correct option is B $1.89\times {10}^{33}$

1st let us calculate the energy of one photon which has a frequency of 800 Hz.

we know that energy of a radiation or photon is given by : E = hν

where h is the planks constant and is equal to 6.625 × 10⁻³⁴ J-sec and given the frequency of photon = ν = 800 Hz

substituting in the energy equation we get :

E = 6.625 × 10⁻³⁴ J-sec × 800 sec⁻¹

⇒ E = 5.3 × 10⁻³¹ Joules.

⇒ The energy of one photon which has frequency 800 Hz is 5.3 × 10⁻³¹ Joules.

given the power of 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 energy of one photon is 5.3 × 10⁻³¹ Joules.

⇒ if energy of one photon is 5.3 × 10⁻³¹ Joules then 1000 joules contains how many photons is what we need to calculate.

⇒ Number of photons which make up an energy of 1000 joules is :

= (1000 joules × 1 photon)/(5.3 × 10⁻³¹ Joules)

⇒ 1.88 × 10³³ photons

⇒ The radio transmitter of 1kw power which operates at a frequency of 800 Hz emits 1.88 × 10³³ photons per second.