Question

What is the wavelength of (a) a photon with energy $\mathbf{\text{1.00eV}}$, (b) an electron with energy $\mathbf{\text{1.00eV}}$, (c) a photon of energy $\mathbf{\text{1.00GeV}}$, and (d) an electron with energy $\mathbf{\text{1.00GeV}}$?

Short answer

(a) The wavelength of photon with energy $1.00\mathrm{eV}$is $1240\text{\hspace{0.17em}nm}$.

(b) The wavelength of electron with energy $1.00\mathrm{eV}$is $1.23\text{\hspace{0.17em}nm}$.

(c) The wavelength of photon with energy $1.00\text{\hspace{0.17em}G}\mathrm{eV}$is $1.24\text{\hspace{0.17em}fm}$.

(d) The wavelength of electron with energy $1.00\text{\hspace{0.17em}G}\mathrm{eV}$is $\text{1.24\hspace{0.17em}fm}$.

Step-by-step solution

The given data:

A photon and an electron with energy $1.00\mathrm{eV}$.

A photon and an electron with energy $1.00\text{\hspace{0.17em}G}\mathrm{eV}$.

Definition of de Broglie wavelength:

The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength.

The momentum of photon is given by,

$p=\frac{E}{c}$ ….. (1)

Here, $E$is its energy and $c$is speed of light having a value $3\times {10}^8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

Its wavelength is,

$\lambda =\frac{h}{p}$ ….. (2)

Here, $p$is momentum of particle and $h$is Plank’s constant having a value $6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$.

So, by substituting equation (1) the wavelength of photon will be,

$\lambda =\frac{hc}{E}$ ….. (3)

Here, the value of constant $hc$ is $1240\text{\hspace{0.17em}eV}\cdot \text{nm}$.

The momentum of electron is given by

$p=\sqrt{2mK}$ ….. (4)

Here, $K$ is kinetic energy and $m$is its mass of electron having a value $9.109\times {10}^{-31}\text{\hspace{0.17em}kg}$.

Substitute $\sqrt{2mK}$ for $p$into equation (2).

$\lambda =\frac{h}{\sqrt{2mK}}$ ….. (5)

(a) Determining the wavelength of photon with energy 1.00 eV:

Energy is given to be $E=1\text{\hspace{0.17em}eV}$.

Write the equation for wavelength as below.

$\lambda =\frac{hc}{E}$

Substitute known values in the above equation.

$\begin{array}{rcl}\lambda & =& \frac{1240\text{\hspace{0.17em}eV}\cdot \text{nm}}{1\text{\hspace{0.17em}eV}}\\ & =& 1240\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength of photon with energy $1.00\mathrm{eV}$is $\text{1240nm}$.

(b) Determining the wavelength of electron with energy 1.00 eV:

The wavelength of electron is defined by,

$\lambda =\frac{h}{\sqrt{2mK}}$

Substitute known values in the above equation, and you have

$\begin{array}{rcl}\lambda & =& \frac{6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{\sqrt{2(9.109\times {10}^{-31}\text{\hspace{0.17em}kg})\left(1.602\times {10}^{-19}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{eV}$}\right.\right)K}}\\ & =& \frac{1.226\times {10}^{-9}\text{\hspace{0.17em}m}\cdot {\text{eV}}^{1/2}}{\sqrt{K}}\\ & =& \frac{1.226\text{\hspace{0.17em}nm}\cdot {\text{eV}}^{1/2}}{\sqrt{K}}\end{array}$

It is given that$K=1\text{\hspace{0.17em}eV}$

Therefore, the wavelength will be,

$\begin{array}{rcl}\lambda & =& \frac{1.226\text{\hspace{0.17em}nm}\cdot {\text{eV}}^{1/2}}{\sqrt{1\text{\hspace{0.17em}eV}}}\\ & =& 1.23\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength electron with energy $1.00\mathrm{eV}$ is $\begin{array}{rcl}& & 1.23\text{\hspace{0.17em}nm}\end{array}$.

(c) Determining the wavelength of photon with energy 1.00 GeV:

Energy is given to be,

$E=1\text{\hspace{0.17em}\hspace{0.17em}GeV}=1\times {10}^9\text{\hspace{0.17em}eV}$

The wavelength of photon is,

$\begin{array}{rcl}\lambda & =& \frac{hc}{E}\\ & =& \frac{1240\text{eV}\cdot \text{nm}}{1\times {10}^9\text{eV}}\\ & =& 1.24\times {10}^{-6}\text{\hspace{0.17em}nm}\\ & =& 1.24\text{fm}\end{array}$

Hence, the wavelength of photon with energy $1.00\text{G}\mathrm{eV}$is $\begin{array}{rcl}& & 1.24\text{fm}\end{array}$.

(d) Determining the wavelength of electron with energy 1.00 GeV:

Here, wavelength of electron can be found using relativity theory.

The momentum $p$ and kinetic energy $K$ are related as,

${\left(pc\right)}^2=K^2+2Km{c}^2$

$pc=\sqrt{K^2+2Km{c}^2}$ ….. (6)

The kinetic energy is given as,

$K=1\text{\hspace{0.17em}\hspace{0.17em}GeV}=1\times {10}^9\text{\hspace{0.17em}eV}$

Putting known values into equation (6) and you get

$\begin{array}{rcl}pc& =& \sqrt{{(1\times {10}^9\text{\hspace{0.17em}eV})}^2+2(1\times {10}^9\text{\hspace{0.17em}eV})(0.511\times {10}^6\text{\hspace{0.17em}eV})}\\ & =& 1\times {10}^9\text{\hspace{0.17em}eV}\end{array}$

So the wavelength is,

$\begin{array}{rcl}\lambda & =& \frac{h}{p}\\ & =& \frac{hc}{pc}\end{array}$

Thus, the wavelength is,

$\begin{array}{rcl}\lambda & =& \frac{1240\text{\hspace{0.17em}eV}\cdot \text{nm}}{1\times {10}^9\text{\hspace{0.17em}eV}}\\ & =& 1.24\times {10}^{-6}\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength of electron with energy $1.00\text{G}\mathrm{eV}$is $\begin{array}{rcl}& & 1.24\times {10}^{-6}\text{\hspace{0.17em}nm}\end{array}$.