Question

A bulb emits light of wavelength 4500$A^o$. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second?

Answer 1

Hint : We have to find how many photons are emitted by the bulb per second .Therefore, we use the formula for the number of photons emitted per second.
$n\; = \;\dfrac{{P \times n \times \lambda }}{{hc}}\;$
Given:
Power of the bulb = 150 watt
Energy Efficiency of the bulb = 8% = $\dfrac{8}{{100}}$
A bulb emits light of wavelength.

Complete step by step solution :
We must know that the each photon has energy of hʋ,
where
h = planck's constant
ʋ = frequency of light
consider, N photons are emitted
then, the total energy radiated is equal to$N{\text{ }} \times h\upsilon$.
by the formula, Average power is given as follows,

$$P = \dfrac{{Energy}}{{time}} \\\ P = \dfrac{E}{t} \\\ P = \dfrac{{Nh\upsilon }}{t} \\\$$

the number of photons emitted per second = $n{\text{ }} = \dfrac{N}{t}$
$n = \dfrac{P}{{h\nu }},$
$but,\nu = \dfrac{c}{\lambda }$
$\therefore n = \dfrac{{P\lambda }}{{hc}}$
as mentioned, all the power is not radiated so, the effective power = rated power x efficiency.
So the number of photons emitted per second

$$n = \dfrac{{P \times eff \times \lambda }}{c} \\\ n\; = \;\dfrac{{P \times n \times \lambda }}{{hc}}\; \\\$$

where η is the efficiency.
As given in question,
P = 150W,
wavelength = $\lambda {\text{ }} = {\text{ }}4500{\text{ }}{A^o} = {\text{ }}4500{\text{ }} \times {\text{ }}{10^{ - 10}}m$
energy efficiency =$n{\text{ }} = {\text{ }}8\% {\text{ }} = {\text{ }}0.08$
Planck’s constant = $h{\text{ }} = {\text{ }}6.6 \times {10^{ - 34}}Js$
Speed of light = $c{\text{ }} = 3 \times {10^8}m/s$
from the above information and putting these values in the formula, we get
number of photons emitted per second,

$$n = \dfrac{{\left( {150{\text{ }}watt{\text{ }} \times 0.08 \times 4500{\text{ }} \times {\text{ }}{{10}^{ - 10}}m} \right)}}{{(6.6 \times {{10}^{ - 34}}Js)\; \times (3 \times {{10}^8}m/s)\;}}\; \\\ n{\text{ }} = {\text{ }}27.2 \times {10^{18\;}} \\\$$

Hence, $27.2 \times {10^{18\;}}$ number of photons will be emitted per second by the bulb.
Note : We must know that we can detect individual photons by several methods. We can use classic photomultiplier tubes to exploit the photoelectric effect: a photon of sufficient energy strikes a metal plate and knocks free an electron, initiating an ever-amplifying avalanche of electrons. And also semiconductor charge-coupled device chips use a similar effect: an incident photon generates a charge on a microscopic capacitor that can be detected.