Electrons accelerated to an energy of $\mathbf{\text{50\hspace{0.17em}GeV}}$have a de Broglie wavelength $\mathit{\lambda }$small enough for them to probe the structure within a target nucleus by scattering from the structure. Assume that the energy is so large that the extreme relativistic relation $\mathit{p}\mathbf{=}\raisebox{1ex}{$\mathbf{E}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{c}$}\right.$between momentum magnitude $\mathit{p}$and energy $\mathit{E}$applies. (In this extreme situation, the kinetic energy of an electron is much greater than its rest energy.)

(a) What is $\mathit{\lambda }$?

(b) If the target nucleus has radius $\mathit{R}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{fm}$, what is the ratio $\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\lambda }$}\right.$?

Electrons accelerated to an energy of $K=50\text{GeV}=50\times {10}^9\text{eV}$.

Target nucleus radius is$R=5\text{\hspace{0.17em}fm}$.

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

Concept used: The greater the potential difference, the greater the kinetic energy and momentum, and smaller is the de Broglie wavelength.

Relativistic formula is,

$\begin{array}{rcl}p& =& \sqrt{{\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)}^2-{m_e}^2{c}^2}\\ & =& \raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\\ & =& \raisebox{1ex}{$K$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\end{array}$

Here, $p$is the momentum, $E$is the energy, $k$is the kinetic energy, $m_e$is mass of electron having a value $9.1\times {10}^{-31}\text{kg}$, $c$is speed of light.

The de Broglie wavelength is,

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

Here, $h$ is Planck’s constant and $p$is momentum.

The value of constant $hc$ is,

$hc=1240\text{eV}\cdot \text{nm}$

Now, de Broglie wavelength is,

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

Substitute known values in the above equation.

role="math" localid="1663136364766" $\begin{array}{rcl}\lambda & =& \frac{1240\text{eV}\cdot \text{nm}}{50\times {10}^9\text{eV}}\\ & =& 2.5\times {10}^{-8}\text{\hspace{0.17em}nm}\end{array}$

Hence, the required de Broglie wavelength is $\begin{array}{rcl}& & 2.5\times {10}^{-8}\text{\hspace{0.17em}nm}\end{array}$.

The wavelength is,

$\begin{array}{rcl}\lambda & =& 2.5\times {10}^{-8}\text{nm}\\ & =& 0.025\text{fm}\end{array}$

Determine the ratio of Target nucleus radius and d Broglie wavelength as below.

$\begin{array}{rcl}\frac{R}{\lambda }& =& \frac{5\text{\hspace{0.17em}fm}}{0.025\text{\hspace{0.17em}fm}}\\ & =& 200\end{array}$

Hence, the required ratio of $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$is $200$.