A 200 watt bulb emits monochromatic light of wavelength 400 nm. Calculate
the number of photons emitted per second by the bulb. (Note: Number of
photons – Power of the bulb /Energy of photons, 200 watt = 200J/s)
Answers
Given :-
◉ A 200 watt bulb emits monochromatic light of wavelength 400 nm
To Find :-
◉ Number of photons emitted per second by the bulb.
Solution :-
We know,
⇒ Power of the Bulb = 200 Watt
Which means, The bulb emits light of 200 J per second.
So, Now let us calculate the energy of one photon of the light emitted by the bulb.
⇒ E = hv
⇒ E = hc/λ
Where,
- c = Speed of Light
- λ = Wavelength
⇒ E = ( 6.6 × 10⁻³⁴ × 3 × 10⁸ ) / ( 400 × 10⁻⁹ )
⇒ E = ( 19.8 × 10⁻²⁶ ) / ( 4 × 10⁻⁷ )
⇒ E = 4.95 × 10⁻¹⁹ Joule ...(1)
Now that we have got the energy of one photon, Let us find the number of photons emitted per second.
⇒ Energy per second = 200 Joules
⇒ n × E = 200
⇒ n = 200 / (4.95 × 10⁻¹⁹ )
⇒ n = 40.40 × 10¹⁹
⇒ n = 4.04 × 10²⁰
Hence, 4.04 × 10²⁰ photons would be emitted per second.
Some Information :-
◉ The energy of a photon is expressed as:
- E = hv , v = frequency
- E = hc/λ , λ = wavelength