Question

A monochromatic light source $6\times 10^{14}Hz$ is emitting energy at the rate $2\times10^{-3}{J}/{s}$. Calculate the number of photons emitted per second by the source.

Answer 1

Hint: To be solved using the Planck energy of the photon equation. This equation helps in calculating the energy of the photon when their frequency $\nu$ or wavelength $\lambda$ is given. Planck's equation: $E=n\cdot h\cdot\nu$

where, $E=$ energy, $n=$number of photons, $\nu=$ the frequency, $h=$ Planck's constant.

Complete solution Step-by-Step:
We know that light has different characteristics such as frequency $\nu$ and wavelength $\lambda$. We also know that, photon energy is the energy carried by a single photon, which is directly proportional to the photon's electromagnetic frequency or is inversely proportional to the wavelength. Thus, the greater the photon’s frequency, the greater its energy. Equivalently, the longer the photon’s wavelength, the lesser its energy. Photon energy can be expressed using any unit of energy. The commonly used units are the electronvolt (eV) and the joule (J). As $1J=6.24 × 10^{18} eV$.

Planck's law helps in calculating the energy of the photon when their frequency $\nu$ or wavelength $\lambda$ is given. Or we can also say that Planck's constant gives the relation between the frequency $\nu$ or the wavelength $\lambda$. of the electromagnetic radiation and the energy of the photon.

The energy of the light is given by the Planck's equation: $E=n\cdot h\cdot\nu$
Then $n$ number of photons $n=\dfrac{E} {h.\nu}$

Given, $\nu=6\times10^{14}Hz$ , $E=2\times10^{-3}{J}/{s}$.
Also, Planck’s constant $h=6.626\times {{10}^{-34}}J/s$

Substituting, we get

$n=\dfrac{2\times10^{-3}{J}/{s}}{6\times10^{14}Hz\times 6.626 \times10^{-34}{J}/{s}}=0.05\times10^{17}$,
The number of photons emitted by the source is 0.05 X 1017

Additional Information:
A German physicist Max Planck in the 1900, gave the theory of black body radiations, he stated that the spectral lines of a hypothetical black body would emit radiations in small, discrete packets called the quanta of energy, and not as continuous radiation as expected.

Note: Be careful with the units and powers used.Photon energy is the energy carried by a single photon, which is directly proportional to the photon's electromagnetic frequency or is inversely proportional to the wavelength. The greater the photon’s frequency, the greater its energy. Equivalently, the longer the photon’s wavelength, the lesser its energy. Photon energy can be expressed using any unit of energy. The commonly used units are the electronvolt (eV) and the joule (J). As $1J=6.24 \times 10^{18} eV$.