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.