91Ó°ÊÓ

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

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

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 $3×{10}^8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

Its wavelength is,

$\mathrm{λ}=\frac{h}{p}$ ….. (2)

Here, $p$is momentum of particle and $h$is Plank’s constant having a value $6.626×{10}^{−34}\text{ J}â‹\dots \text{s}$.

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

$\mathrm{λ}=\frac{hc}{E}$ ….. (3)

Here, the value of constant $hc$ is $1240\text{ e³Õ}â‹\dots \text{nm}$.

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 $9.109×{10}^{−31}\text{ k²µ}$.

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

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

Energy is given to be $E=1\text{ e³Õ}$.

Write the equation for wavelength as below.

$\mathrm{λ}=\frac{hc}{E}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{λ}& =& \frac{1240\text{ e³Õ}â‹\dots \text{nm}}{1\text{ e³Õ}}\\ & =& 1240\text{ n³¾}\end{array}$

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

The wavelength of electron is defined by,

$\mathrm{λ}=\frac{h}{\sqrt{2mK}}$

Substitute known values in the above equation, and you have

$\begin{array}{rcl}\mathrm{λ}& =& \frac{6.626×{10}^{−34}\text{ J}â‹\dots \text{s}}{\sqrt{2(9.109×{10}^{−31}\text{ k²µ})\left(1.602×{10}^{−19}\text{ }\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{eV}$}\right.\right)K}}\\ & =& \frac{1.226×{10}^{−9}\text{″¾}â‹\dots {\text{eV}}^{1/2}}{\sqrt{K}}\\ & =& \frac{1.226\text{ n³¾}â‹\dots {\text{eV}}^{1/2}}{\sqrt{K}}\end{array}$

It is given that$K=1\text{ e³Õ}$

Therefore, the wavelength will be,

$\begin{array}{rcl}\mathrm{λ}& =& \frac{1.226\text{ n³¾}â‹\dots {\text{eV}}^{1/2}}{\sqrt{1\text{ e³Õ}}}\\ & =& 1.23\text{ n³¾}\end{array}$

Hence, the wavelength electron with energy $1.00\mathrm{eV}$ is $\begin{array}{rcl}& & 1.23\text{ n³¾}\end{array}$.

Energy is given to be,

$E=1\text{ â¶Ä‰G±ð³Õ}=1×{10}^9\text{ e³Õ}$

The wavelength of photon is,

$\begin{array}{rcl}\mathrm{λ}& =& \frac{hc}{E}\\ & =& \frac{1240\text{eV}â‹\dots \text{nm}}{1×{10}^9\text{eV}}\\ & =& 1.24×{10}^{−6}\text{ n³¾}\\ & =& 1.24\text{fm}\end{array}$

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

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,

$K=1\text{ â¶Ä‰G±ð³Õ}=1×{10}^9\text{ e³Õ}$

$\begin{array}{rcl}pc& =& \sqrt{{(1×{10}^9\text{ e³Õ})}^2+2(1×{10}^9\text{ e³Õ})(0.511×{10}^6\text{ e³Õ})}\\ & =& 1×{10}^9\text{ e³Õ}\end{array}$

So the wavelength is,

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

Thus, the wavelength is,

$\begin{array}{rcl}\mathrm{λ}& =& \frac{1240\text{ e³Õ}â‹\dots \text{nm}}{1×{10}^9\text{ e³Õ}}\\ & =& 1.24×{10}^{−6}\text{ n³¾}\end{array}$

Hence, the wavelength of electron with energy $1.00\text{G}\mathrm{eV}$is $\begin{array}{rcl}& & 1.24×{10}^{−6}\text{ n³¾}\end{array}$.