If the de Broglie wavelength of a proton is $\mathbf{\text{100fm}}$, (a) what is the speed of the proton and (b) through what electric potential would the proton have to be accelerated to acquire this speed?

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

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

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, $m=1.6705\times {10}^{-27}\text{kg}$

The charge, $e=1.60\times {10}^{-19}\text{C}$

Plank’s constant, $h=6.26\times {10}^{-34}\text{J}\cdot \text{s}$

The wavelength, $\lambda =100\text{fm}=1.0\times {10}^{-13}\text{m}$

Since the wavelength is,

$\lambda =\frac{h}{mv}$

Rearrange the above equation for velocity as below.

$v=\frac{h}{m\lambda }$

Substitute known values in the above equation.

$\begin{array}{rcl}v& =& \frac{6.26\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{(1.6705\times {10}^{-27}\text{\hspace{0.17em}kg})(1.0\times {10}^{-13}\text{\hspace{0.17em}m})}\\ & =& 3.96\times {10}^6\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Hence, the speed of the proton is $\begin{array}{rcl}& & 3.96\times {10}^6\raisebox{1ex}{$\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" $\begin{array}{rcl}\text{V}& =& \frac{(1.6705\times {10}^{-27}\text{\hspace{0.17em}kg})(3.96\times {10}^6\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)}{2(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}\\ & =& 8.18\times {10}^4\text{V}\left(\frac{\text{1\hspace{0.17em}kV}}{{\text{10}}^{\text{3}}\text{\hspace{0.17em}V}}\right)\\ & =& 81.8\text{kV}\end{array}$

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