91Ó°ÊÓ

Given that the de Broglie wavelength of a proton is $100\text{fm}$.

The wavelength that is associated with an object in relation to its momentum and mass is known as the de Broglie wavelength. The de Broglie wavelength of a particle is usually inversely proportional to its strength.

The de Broglie wavelength is given by

$$\begin{gathered} \mathrm{λ}=\frac{h}{p} \\ =\frac{h}{mv} \end{gathered}$$

Here, $p$is momentum, $h$is Plank’s constant, $m$is the mass of the particle, and $c$is the speed of the particle.

Also, the energy is

$\begin{array}{rcl}K& =& \frac{1}{2}m{v}^2\\ & =& eV\end{array}$

Here, $V$is electric potential, $e$ is the charge on the electron, $m$ is the mass of the particle, $v$and is the speed of the particle.

Consider the known data as below.

Mass of proton, $m=1.6705×{10}^{−27}\text{kg}$

The charge, $e=1.60×{10}^{−19}\text{C}$

Plank’s constant, $h=6.26×{10}^{−34}\text{J}â‹\dots \text{s}$

The wavelength, $\mathrm{λ}=100\text{fm}=1.0×{10}^{−13}\text{m}$

Since the wavelength is,

$\mathrm{λ}=\frac{h}{mv}$

Rearrange the above equation for velocity as below.

$v=\frac{h}{m\mathrm{λ}}$

Substitute known values in the above equation.

$\begin{array}{rcl}v& =& \frac{6.26×{10}^{−34}\text{ J}â‹\dots \text{s}}{(1.6705×{10}^{−27}\text{ k²µ})(1.0×{10}^{−13}\text{″¾})}\\ & =& 3.96×{10}^6\text{ }\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Hence, the speed of the proton is $\begin{array}{rcl}& & 3.96×{10}^6\text{ }\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$.

Since

$eV=\frac{\text{1}}{\text{2}}m{v}^2$

Rearrange the above equation for potential.

$\begin{array}{rcl}V& =& \frac{m{v}^2}{2e}\end{array}$

Substitute known values in the above equation.

role="math" localid="1663148728982" $\begin{array}{rcl}\text{V}& =& \frac{(1.6705×{10}^{−27}\text{ k²µ})(3.96×{10}^6\text{ }\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)}{2(1.6×{10}^{−19}\text{ C})}\\ & =& 8.18×{10}^4\text{V}\left(\frac{\text{1 k³Õ}}{{\text{10}}^{\text{3}}\text{ V}}\right)\\ & =& 81.8\text{kV}\end{array}$

Hence, the electric potential would the proton has to be accelerated to acquire this speed is $\begin{array}{rcl}& & 81.8\text{kV}\end{array}$.