Chapter 27.2, Problem 25PP

Solution

To Calculate: The de Broglie wavelength and speed of an electron.

The De-Broglie wavelength is = $7.77\times {10}^{-11}\text{m}$

The speed of an electron is = $9.376\times {10}^6\text{m}\cdot {\text{s}}^{\text{-1}}$

Given data: The electron accelerated with a potential difference $\left(V\right)=250\text{V}$

Formula used:

  $KE=\frac{1}{2}m{v}^2=eV$

Where,

KE = kinetic energy

m =mass

e = charge of electron

v = velocity

V = potential difference

Re-arrange the equation,

  $v={\left(\frac{2eV}{m}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the required formula for velocity,

De-Broglie wavelength is:

  $\lambda =\frac{h}{mv}$

h =Planck’s constant

  $\lambda$ = wavelength

Calculation:

Mass of electron is = $9.1\times {10}^{-31}\text{kg}$

Charge of electron is = $1.6\times {10}^{-19}\text{C}$

Substitute this in above formula also using:

  $\begin{array}{l}v=\sqrt{\frac{2eV}{m}}\\ v=\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 250}{9.1\times {10}^{-31}}}\\ v=9.376\times {10}^6\text{m}\cdot {\text{s}}^{\text{-1}}\end{array}$

Therefore, velocity is = $9.376\times {10}^6\text{m}\cdot {\text{s}}^{\text{-1}}$

Planck’s constant (h)= $6.6\times {10}^{-34}\text{J}\cdot {\text{s}}^{\text{-1}}$

De-Broglie wavelength, $\lambda =\frac{h}{mv}$

  $\begin{array}{l}\lambda =\frac{6.6\times {10}^{-34}}{9.1\times {10}^{-31}\times 9.376\times {10}^6}\\ \lambda =7.77\times {10}^{-11}\text{m}\end{array}$

Conclusion:

The de Broglie wavelength and speed of electron are $7.77\times {10}^{-11}\text{m}$ and $9.376\times {10}^6\text{m}\cdot {\text{s}}^{-1}$ respectively.