Question

Calculate the Compton wavelength for
(a) an electron and
(b) a proton.
What is the photon energy for an electromagnetic wave with wavelength equal to Compton wavelength of
(c) the electron and
(d) the proton?

Answer 1

In order to solve this question we need to understand Compton scattering. Compton scattering is a phenomenon in which an electron at rest strikes by light of wavelength, which gets scattered after striking electrons; the wavelength of emitted electrons is found to be greater than incident light wavelength. During Compton scattering, light shows a particle-like nature as light is made up of photons having energy $hv$ . Similarly photoelectric effect, blackbody radiation and pair production are the phenomena in which light shows its particle nature, also interference, diffraction etc. are the phenomena in which light shows wave nature.

Complete step by step answer:
(a) Consider a light or photon having wavelength $\lambda$. Let the mass of the electron be ${m_e}$. Let the photon be reflected after striking at an angle $\phi$. So increase in wavelength during Compton scattering is given by,
$\Delta \lambda = {\lambda _0}(1 - \cos \phi )$
Here, $\Delta \lambda$ is an increase in wavelength and ${\lambda _0}$ is Compton wavelength.

Compton wavelength is defined as; ${\lambda _0} = \dfrac{h}{{{m_e}c}}$
Since, $h$ is known as Planck’s constant and it is equal to, $h = 6.626 \times {10^{ - 34}}Js$ and “c” is light velocity, $c = 3 \times {10^8}m{\sec ^{ - 1}}$
And mass of electron is, ${m_e} = 9.1 \times {10^{ - 31}}kg$
So Compton wavelength is,
${\lambda _0} = \dfrac{{6.626 \times {{10}^{ - 34}}Js}}{{(3 \times {{10}^8})(9.1 \times {{10}^{ - 31}}kg)}}$
$\therefore {\lambda _0} = 0.243 \times {10^{ - 11}}m$

(b) Let the mass of proton be ${m_p}$ and it is equal to ${m_p} = 1.67 \times {10^{ - 28}}kg$
So Compton wavelength is, $\lambda {'_0} = \dfrac{h}{{{m_p}c}}$
So, Compton wavelength is,
$\lambda {'_0} = \dfrac{{6.626 \times {{10}^{ - 34}}Js}}{{(3 \times {{10}^8})(1.67 \times {{10}^{ - 28}}kg)}}$
$\Rightarrow \lambda {'_0} = 1.32 \times {10^{ - 14}}m$
We know the energy of a photon is equal to, $E = \dfrac{{hc}}{\lambda }$

(c) For electron, Compton wavelength is ${\lambda _0} = 0.243 \times {10^{ - 11}}m$
So, the energy of a photon is, $E = \dfrac{{hc}}{{{\lambda _0}}}$
Putting values we get,
$E = \dfrac{{(6.626 \times {{10}^{ - 34}}Js)(3 \times {{10}^8}m{{\sec }^{ - 1}})}}{{0.243 \times {{10}^{ - 11}}m}}$
$\therefore E = 81.8 \times {10^{ - 15}}J$

(d) For protons, Compton wavelength is $\lambda {'_0} = 1.32 \times {10^{ - 14}}m$.
So, the energy of a photon is, $E = \dfrac{{hc}}{{\lambda {'_0}}}$
Putting values we get,
$E = \dfrac{{(6.626 \times {{10}^{ - 34}}Js)(3 \times {{10}^8}m{{\sec }^{ - 1}})}}{{1.32 \times {{10}^{ - 14}}m}}$
$\therefore E = 15.06 \times {10^{ - 12}}J$

Note: It should be remembered that Compton scattering could not be explained by wave nature of light as in Compton scattering there is frequency change which is restricted by wave motion because frequency is considered to be property of electromagnetic waves and cannot be altered by change in direction. So Compton wavelength is explained by the particle nature of light.