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Expert-verifiedUsing the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as ${\mathit{\lambda}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{226}}{\mathbf{/}}\sqrt{\mathbf{K}}$, in which K is the electron’s kinetic energy in electron-volts.
The required expression for the de Broglie wavelength is, $\frac{1.226\times {10}^{-9}{\text{m.eV}}^{\frac{\text{1}}{\text{2}}}}{\sqrt{K}}$ .
The classical equation of momentum is, $p=mv$
The kinetic energy is, $K=\frac{1}{2}m{v}^{2}$
The expression to calculate the kinetic energy of the electron is given as follows.
$K=\frac{1}{2}m{v}^{2}$ …(i)
Here, m is the mass of the electron and v is the velocity of electron.
The expression to calculate the de Broglie wavelength is given as follows.
$\lambda =\frac{h}{p}$ …(ii)
Here, h is the plank’s constant and $p$ is the momentum.
Calculate the kinetic energy of the electron,
Substitute $\rho /m$ for v into equation (i).
$\begin{array}{rcl}K& =& \frac{1}{2}m{\left(\frac{p}{m}\right)}^{2}\\ K& =& \frac{1}{2}\frac{{p}^{2}}{m}\\ {p}^{2}& =& 2Km\\ p& =& \sqrt{2Km}\end{array}$
Calculate the de Broglie wavelength.
Substitute$\sqrt{2Km}$ for $p$ into equation (ii).
$\lambda =\frac{h}{\sqrt{2Km}}$
Substitute $6.62\times {10}^{-34}\text{Js}$ for h and $9.11\times {10}^{-31}\text{kg}$for m into above equation.
$\begin{array}{rcl}\lambda & =& \frac{6.62\times {10}^{-34}\mathrm{Js}}{\sqrt{2\times \left(9.11\times {10}^{-31}\mathrm{kg}\right)\times \left(1.602\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)\times K}}\\ & =& \frac{6.62\times {10}^{-34}\mathrm{Js}}{\sqrt{29.1884\times {10}^{-50}\mathrm{kgJ}/\mathrm{eV}}}\frac{1}{\sqrt{K}}\\ & =& \frac{1.226\times {10}^{-9}{\mathrm{meV}}^{\frac{1}{2}}}{\sqrt{K}}\end{array}$
Hence the required expression for the de Broglie wavelength is, $\begin{array}{rcl}& & \frac{1.226\times {10}^{-9}{\mathrm{meV}}^{\frac{1}{2}}}{\sqrt{K}}\end{array}$.The table gives relative values for three situations for the barrier tunneling experiment of Figs. 38-16 and 38-17. Rank the situations according to the probability of the electron tunneling through the barrier, greatest first.
| Electron Energy | Barrier Height | Barrier Thickness |
(a) | E | 5E | L |
(b) | E | 17E | L/2 |
(c) | E | 2E | 2L |
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