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.
$$n={\displaystyle \frac{P\times n\times \lambda }{hc}}$$
Given:
Power of the bulb = 150 watt
Energy Efficiency of the bulb = 8% = ${\displaystyle \frac{8}{100}}$
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$$N\times h\upsilon$$.
by the formula, Average power is given as follows,
 $$\begin{gathered} P={\displaystyle \frac{Energy}{time}} \\ P={\displaystyle \frac{E}{t}} \\ P={\displaystyle \frac{Nh\upsilon }{t}} \end{gathered}$$
the number of photons emitted per second = $n={\displaystyle \frac{N}{t}}$
$$n={\displaystyle \frac{P}{h\nu }},$$
$$but,\nu ={\displaystyle \frac{c}{\lambda }}$$
$$\therefore n={\displaystyle \frac{P\lambda }{hc}}$$
as mentioned, all the power is not radiated so, the effective power = rated power x efficiency.
So the number of photons emitted per second
$$\begin{gathered} n={\displaystyle \frac{P\times eff\times \lambda }{c}} \\ n={\displaystyle \frac{P\times n\times \lambda }{hc}} \end{gathered}$$
 where η is the efficiency.
As given in question,
P = 150W,
wavelength = $$\lambda =4500A^o=4500\times {10}^{-10}m$$
energy efficiency =$$n=8\mathrm{\%}=0.08$$
Planck’s constant = $$h=6.6\times {10}^{-34}Js$$
Speed of light = $$c=3\times {10}^{8}m/s$$
from the above information and putting these values in the formula, we get
number of photons emitted per second,
$$\begin{gathered} n={\displaystyle \frac{\left(150watt\times 0.08\times 4500\times {10}^{-10}m\right)}{(6.6\times {10}^{-34}Js)\times (3\times {10}^{8}m/s)}} \\ n=27.2\times {10}^{18} \end{gathered}$$
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.