Question

What is the number of photons emitted in $$10$$ hours by a $$60W$$ sodium lamp? $$(\lambda$$ of photon $$=6000A^{\circ })$$

Answer 1

Hint :A form of an elementary particle is the photon. It is the force carrier for the electromagnetic force and the quantum of the electromagnetic field, which includes electromagnetic radiation such as light and radio waves. Since photons have no mass, they still travel at the speed of light in a vacuum.

Complete Step By Step Answer:
Given that, Power $$P=60W$$ and Time $$T=10hours=10\times 3600\sec =3.6\times {10}^4\sec$$
We know that, $$P={\displaystyle \frac{E}{T}}$$ , where $$E$$ is Energy.
That is, $$E=P\times T$$
Therefore, $$E=60W\times 3.6\times {10}^4\sec =21.6\times {10}^{5}J$$
Hence, $$E=2.16\times {10}^{6}J$$
Now using Planck’s quantum theory, we are going to find out the energy of one photon.
According to Planck’s quantum theory,
Energy of the radiation, $$E=h\nu$$ where $$h$$ is Planck’s constant and $$\nu$$ is the frequency of radiation.
We are familiar with $$\nu ={\displaystyle \frac{c}{\lambda }}$$ where $$c$$ is the speed of light and $$\lambda$$ is the wavelength of the light.
So, after substituting the formula of $$\nu$$ , we got $$E={\displaystyle \frac{hc}{\lambda }}$$
Given that, $$\lambda =6000A^{\circ }=6\times {10}^{-7}m$$
We know that $$c=3\times {10}^8$$ m/s and $$h=6.63\times {10}^{-34}$$ kg/s
Therefore, Energy of one photon $$E={\displaystyle \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{6\times {10}^{-7}}}={\displaystyle \frac{19.89}{6\times {10}^{-7}}}\times {10}^{-19}J$$
Hence, we can find that the number of photons $$=$$ Energy / Energy of one photon
$$=2.16\times {10}^6÷{\displaystyle \frac{19.89}{6\times {10}^{-7}}}\times {10}^{-19}$$
$$=2.16\times {\displaystyle \frac{6}{19.89}}\times {10}^{25}$$
$$=0.64\times {10}^{25}$$
So, the number of photons emitted in $$10$$ hours by a $$60W$$ sodium lamp is $$6.4\times {10}^{24}$$

Note :
We should remember the formulas, $$P={\displaystyle \frac{E}{T}}$$ and $$E={\displaystyle \frac{hc}{\lambda }}$$ . Don’t forget that the speed of light $$c=3\times {10}^8$$ m/s, which we want to use in many solutions.
According to Planck’s Quantum Theory,
Only discrete amounts of energy can be emitted or absorbed by different atoms and molecules. Quantum energy is the smallest amount of energy that can be released or consumed in the form of electromagnetic radiation.
The frequency of the emission is directly proportional to the energy of the emissions generated or released.