Question

Infrared lamps are used in restaurants to keep the food warm. The infrared radiation is strongly absorbed by water, raising its temperature and that of the food. If the wavelength of infrared radiation is assumed to be $1500\;nm$, then the number of photons per second of infrared radiation produced by an infrared lamp, that consumes energy at the rate of $100\; W$ and is $12\%$ efficient only, is $(x \times 10^{19})$. The value of x is:
(Given h = $6.625 \times 10^{-34}\;J.s$)

Answer 1

Hint: Recall that infrared lamps convert electrical energy to light energy of IR photons. First, find the actual amount of power consumed by the lamp for this conversion given its efficiency. This will be the amount of electrical energy $E_{consumed}$ that the lamp consumes to produce photons.
Then, by using the relation between the Planck’s constant, speed of light, and the wavelength of the IR photon, determine the energy $E_{photon}$ of a single photon. Thus, n photons will possess a total energy of $nE_{photons}$.
To this end, equate the total photons energy and the electrical energy consumed to determine the number of photons.

Formula Used:
Energy possessed by a photon $E_{photon} = \dfrac{hc}{\lambda}$, where h is the Planck’s constant = $(6.625 \times 10^{-34}\;J.s)$, c is the velocity of light(photon)= $(3 \times 10^8\;ms^{-1})$ and $\lambda$ is the wavelength of the photon.

Complete step-by-step solution:
Let the total number of photons be $n$.
Now, the rate of consumption of energy, which is nothing but the power is given to be $P=100\; W = 100\; Js^{-1}$, which means that the energy consumed by the lamp in one second $E = 100\; J$, since $E=Pt$ and $t=1 second$
Given that the lamp utilizes only $12\%$ of this energy to produce IR photons, the energy it actually consumes in one second is
$E_{consumed} = \dfrac{12}{100} \times 100 = 12\;J$
Therefore, the lamp utilizes $12\; J$ to produce $n$ IR photons in 1 second.
Now, the energy possessed by a photon is given by the relation
$E_{photon} = \dfrac{hc}{\lambda}$, where, Planck’s constant $h =6.625 \times 10^{-34}\;J.s$, velocity of photon $c =3 \times 10^8\;ms^{-1}$ and wavelength of IR photon $\lambda = 1500 \times 10^{-9}\;m$
$\Rightarrow E_{photon} = \dfrac{(6.625 \times 10^{-34}) \times (3 \times 10^8)}{ 1500 \times 10^{-9}}= \dfrac{19.875 \times 10^{-26}}{1500 \times 10^{-9}} = 1.325 \times 10^{-19}\; J$
Now, if n IR photons were produced, the total energy would be $nE_{photon}$. But this energy comes from the energy consumed by the lamp for this conversion
$\Rightarrow nE_{photon} = 12 \Rightarrow n =\dfrac{12}{E_{photon}} = \dfrac{12}{1.325 \times 10^{-19}} = 9.06 \times 10^{19}\;photons $
Therefore, from the question, the number of photons is given as $(x \times 10^{19})$.
Therefore, $x \approx 9$

Note: Remember to consider only the actual power consumed by the lamp which you can obtain by its efficiency. Additionally, we were required to find the number of photons “per second”, which is why we had to obtain energy consumed in one second to begin with. Therefore, do not forget to maintain the consistency of the units, and their powers throughout the problem as any discrepancies will result in incorrect results.