Question

What is the energy of a photon of red light that has a frequency of $4.48 \times {10^{14}}Hz$ ?

Answer 1

In order to solve this question we need to understand the definition of energy of photons which states that light is made up of packets of photons that possess a certain definite amount of energy and this energy is discrete. Also when light collides with any particle energy in the form of photon is transferred to that particle, this is the particle nature of light.

Complete step by step answer:
We know that, Energy of $1$ photon carries is equal to $h\nu$
Where $h = 6.626 \times {10^{ - 34}}Js$ is known as Planck’s constant and $\nu$ is frequency of light used and it SI unit is $Hz$ or ${\sec ^{ - 1}}$
Now, according to the question
Frequency of photon of red light is given as $v = 4.48 \times {10^{14}}Hz$
So Energy of one photon of red light is $E = hv$
$E = (6.626 \times {10^{ - 34}}) \times (4.48 \times {10^{14}})J$
$\Rightarrow E = 29.684 \times {10^{ - 20}}J$
Also this energy could be expressed in eV which is read as electron volt, so Energy of red light in eV is,
$E = \dfrac{{29.684 \times {{10}^{ - 20}}}}{{1.67 \times {{10}^{ - 19}}}}$
$\therefore E = 1.78\,eV$

Hence, the energy of a photon is $E = 1.78\,eV$.

Note: It should be remembered that here light is considered to me made up of photons instead of waves. Also red light has the lowest frequency because of that its refractive index is lowered and that is the only reason why red light scatters most in the sky.However photon shape is still unknown and is considered as lumps of energy but for sake of simplicity it is considered to be of spherical shape in nature. Also due to the given value of frequency wavelength of light could also be determined. The charge on an electron has a magnitude of $1.67 \times {10^{ - 19}}C$.