91Ó°ÊÓ

Uniform electric potential,$\mathrm{V}=-200\mathrm{V}$

The total energy of the electron,$\mathrm{E}=500\mathrm{eV}$

The total energy of the electron is the sum of its kinetic energy and potential energy and when a particle's potential energy changes at a boundary, it can reflect even though it would not normally reflect under classical theory.

The potential energy of a charged particle can be given as,

$\mathrm{U}=\mathrm{qV}$

Where, $\mathrm{q}$is charge and $\mathrm{V}$is electric potential.

Therefore, the potential energy of an electron can be given as

${\mathrm{U}}_{\mathrm{e}}=-\mathrm{eV}$

Substituting the values of potential

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{e}}& =& -\mathrm{e}\left(-200\mathrm{V}\right)\\ & =& 200\mathrm{eV}\end{array}$

As known the total energy is the sum of kinetic energy and potential energy.

$\mathrm{E}=\mathrm{KE}+\mathrm{U}$

Therefore, the kinetic energy of the electron can be given as

$\mathrm{KE}=\mathrm{E}-\mathrm{U}$

Now substituting the known values in the above equation.

$\mathrm{KE}=500\mathrm{eV}-200\mathrm{eV}=300\mathrm{eV}$

Hence, the kinetic energy of the electron is $300\mathrm{eV}$.

As known the kinetic energy of the particle can be given as,

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Where, $\mathrm{m}$is mass and v is velocity

And the momentum of the particle is

$\begin{array}{l}\mathrm{p}=\mathrm{mv}\\ \mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}}\end{array}$

Now substituting the velocity into the formula of kinetic energy

$\mathrm{KE}=\frac{1}{2}\mathrm{m}\left(\frac{\mathrm{p}}{\mathrm{m}}\right)=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$

$\mathrm{p}=\sqrt{2\mathrm{mKE}}$

$\begin{array}{rcl}\mathrm{p}& =& \sqrt{2×9.1×{10}^{-31}\mathrm{kg}×300\mathrm{eV}}\\ & =& \sqrt{2×9.1×{10}^{-31}\mathrm{kg}×300\mathrm{eV}×\left(1.6×{10}^{-16}\mathrm{J}/\mathrm{eV}\right)}\\ & =& 9.35×{10}^{-24}\mathrm{kgm}/\mathrm{s}\end{array}$

Hence the momentum of the electron is $9.35×{10}^{24}\mathrm{kgm}/\mathrm{s}$

The kinetic energy of the particle can be given as

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Where, m is mass and v is speed

Therefore,

role="math" localid="1663136772867" $\begin{array}{rcl}\mathrm{v}& =& \sqrt{\frac{2\mathrm{KE}}{\mathrm{m}}}\\ & =& \sqrt{\frac{2×300\mathrm{eV}×1.6×{10}^{-16}\mathrm{JeV}}{9.1×{10}^{-31}\mathrm{kg}}}\\ & =& 1.03×{10}^7\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed of the electron is $1.03×{10}^7\mathrm{m}/\mathrm{s}$.

The de Broglie wavelength can be given as

$\mathrm{λ}=\frac{\mathrm{h}}{\mathrm{p}}$

Where, h is plank constant and p is momentum.

Substituting the values

$\begin{array}{rcl}\mathrm{λ}& =& \frac{6.634×{10}^{-34}\mathrm{J}·\mathrm{s}}{9.35×{10}^{-24}\mathrm{kg}·{\mathrm{ms}}^{-1}}\\ & =& 7.08×{10}^{-11}\mathrm{m}\end{array}$

Therefore, the de Broglie wavelength of the electron is $7.08×{10}^{-11}\mathrm{m}$.

The angular wave number can be given as,

$\mathrm{k}=\frac{2\mathrm{π}}{\mathrm{λ}}$

Where, $\mathrm{λ}$is de Broglie’s wavelength.

$\begin{array}{rcl}\mathrm{k}& =& \frac{2\mathrm{π}}{7.08×{10}^{-11}\mathrm{m}}\\ & =& 8.87×{10}^{10}{\mathrm{m}}^{-1}\end{array}$

Hence, the angular wave number of the particle is $8.87×{10}^{-10}{\mathrm{m}}^{-1}$.