Question

Calculate the Compton wavelength for

(a) an electron and

(b) a proton. What is the photon energy for an electromagnetic wave with a wavelength equal to the Compton wavelength of

(c) the electron and

(d) the proton.

Short answer

Thus, (a) the Compton wavelength for an electron is$2.43\text{\hspace{0.17em}pm}$.

(b) The Compton wavelength for a proton is $1.32\text{\hspace{0.17em}fm}$.

(c) The energy for the electron is $0.511\text{\hspace{0.17em}MeV}$.

(d) The energy for the proton is$939\text{\hspace{0.17em}MeV}$.

Step-by-step solution

The Compton wavelength for an electron.

(a)

The mass of an electron is $m=9.109\times {10}^{-31}\text{kg}$, its Compton wavelength is solved as follows:

$\begin{array}{c}{\lambda }_C=\frac{h}{mc}\\ =\frac{6.626\times {10}^{-34}\text{J.s}}{(9.109\times {10}^{-31}\text{kg})(2.998\times {10}^8\text{m/s})}\\ =2.426\times {10}^{-12}\text{m}\\ =2.43\text{\hspace{0.17em}pm}\end{array}$

Hence, the Compton wavelength for an electron is$2.43\text{\hspace{0.17em}pm}$ .

The Compton wavelength for a proton.

(b)

The mass of a proton is$m=1.673\times {10}^{-27}\text{kg}$, its Compton wavelength is solved as follows:

$\begin{array}{c}{\lambda }_C=\frac{h}{mc}\\ =\frac{6.626\times {10}^{-34}\text{J.s}}{(1.673\times {10}^{-27}\text{kg})(2.998\times {10}^8\text{m/s})}\\ =1.321\times {10}^{-15}\text{m}\\ =1.32\text{\hspace{0.17em}fm}\end{array}$

Hence, the Compton wavelength for a proton is $1.32\text{\hspace{0.17em}fm}$.

 Step 3: The photon energy of an electromagnetic wave in a wavelength equal to the wavelength of an electron.

(c)

Let $hc=1240\text{\hspace{0.17em}eV}\cdot \text{nm}$then it gives,

$\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{\lambda }\end{array}$

Here, E is the energy and $\lambda$is the wavelength.

Thus, the energy for the electron is;

$\begin{array}{c}E=\frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{2.426\times {10}^{-3}\text{nm}}\\ =5.11\times {10}^{5}eV\\ =0.511\text{\hspace{0.17em}MeV}\end{array}$

Hence, the energy for the electron is$0.511\text{\hspace{0.17em}MeV}$ .

The photon energy of an electromagnetic wave in a wavelength equal to the wavelength of a proton

(d)

Let $hc=1240\text{\hspace{0.17em}eV}\cdot \text{nm}$then it gives,

$\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{\lambda }\end{array}$

Here, E is the energy and $\lambda$is the wavelength.

Thus, the energy for the proton is;

$\begin{array}{c}E=\frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{1.321\times {10}^{-6}\text{\hspace{0.17em}nm}}\\ =9.39\times {10}^{8}eV\\ =939\text{\hspace{0.17em}MeV}\end{array}$

Hence, the energy for the proton is$939\text{\hspace{0.17em}MeV}$ .