Question

What is the wavelength of a photon of energy 1 MeV?
(A) $1.2$ m
(B) $1.24$m
(C) $1.24 \times {10^{ - 2}}{A^\circ }$
(D) $1.24{A^\circ }$

Answer 1

Hint
A photon is also known as the basic unit of light. It behaves both like a wave and a particle and carries energy. This energy is inversely proportional to the photon’s wavelength.
Formula used: $E = \dfrac{{hc}}{\lambda }$, where E is the energy of the photon in eV, h is the Planck’s constant, c is the speed of light and $\lambda$ is the wavelength of the photon in ${A^\circ }$.

Complete step by step answer
In this question, we are provided with a photon of a definite energy and are asked to find its wavelength. The data provided to us is:
Energy of the photon $E = 1{\text{MeV}} = {10^6}eV$
We know that the energy of a photon is inversely dependent on its wavelength as:
$E = \dfrac{{hc}}{\lambda }$
For the wavelength, we cross multiply to get:
$\lambda = \dfrac{{hc}}{E}$
We know that the product of Planck’s constant h and the speed of light c is given as 12400 when the units of energy are in electron Volts and that of wavelength are in Angstrom units.
So, putting these values in the above equation gives us:
$\lambda = \dfrac{{12400}}{{{{10}^6}}} = 1.24 \times {10^{ - 2}}{A^\circ }$
Hence, the correct answer is option (C).

Note
The unit of energy electron Volt is commonly used at subatomic levels to specify small energy units better. It is the amount of energy gained by the charge of a single electron moved across an electric potential difference of one volt. It is equivalent to $1.6 \times {10^{ - 19}}$ J, and makes the calculations at quantum level easier.