Question

The energy of a photon of visible light with a maximum wavelength in $eV$ is
A) $1$
B) $1.6$
C) $3.2$
D) $7$

Answer 1

First of all we have to separate the given quantities and convert them into S.I units if they aren’t. Now after that, we have to define the equation of the energy in terms of the Planck's constant and the wavelength of the given light,
$E = \dfrac{{{h_c}}}{\lambda }$.
Here we can see that we have all the quantities required for the equation, so in the next step, we just have to substitute the values to obtain the answer.

Complete step by step answer:
A photon is a particle that represents the quantum of light or any other electromagnetic radiation in general. Therefore almost all the properties of the light along with the speed and the frequency applied to the photon.
Given,
The maximum wavelength of the visible light is $7800{A^0}$
Now we have to perform a step by step process to obtain the answer.
Step 1: To obtain the energy when the wavelength is mentioned we apply the following equation
Now, the energy of the red light would be therefore given by the equation:
$E = \dfrac{{{h_c}}}{\lambda }$
Where, ${h_c} = 12400{A^0}$ (planck's constant) and $\lambda = 7800{A^0}$ ( wavelength of the red light)
Step 2: Now we just have to substitute the values in the equation and solve it in order to simplify and obtain the answer. So, it is
$\Rightarrow E = \dfrac{{12400}}{{7800}}V{A^0}$
$\Rightarrow E = 1.6eV$

$\therefore$ The energy of a photon of visible light with a maximum wavelength in $eV$ is 1.6eV. Hence the correct option is B. $1.6$ eV.

Note:
Photons are considered as the smallest unit of light/electromagnetic energy. They are generally regarded as the particles with zero mass and relatively null electric charge. In many cases the amount of energy a photon has produced makes it behave more like a wave than a particle. This is called the “wave-particle duality” of light.