Blog: A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the...

Hello!

Again, we have a deal with the uncertainty principle for a position and a momentum. Mathematically it is expressed as

$\displaystyle\delta{p}\cdot\delta{x}\geq\frac{{h}}{{{4}\pi}},$

where $\displaystyle\delta{p}={m}\cdot\delta{v}$ is an uncertainty in a momentum and $\displaystyle\delta{x}$ is the uncertainty in a position.

Also we know that the kinetic energy of an electron is equal to $\displaystyle\frac{{{m}_{{{e}{l}}}{{v}}^{{2}}}}{{2}},$ and it is given that the speed is the same as the...

Hello!

Again, we have a deal with the uncertainty principle for a position and a momentum. Mathematically it is expressed as

$\displaystyle\delta{p}\cdot\delta{x}\geq\frac{{h}}{{{4}\pi}},$

where $\displaystyle\delta{p}={m}\cdot\delta{v}$ is an uncertainty in a momentum and $\displaystyle\delta{x}$ is the uncertainty in a position.

Also we know that the kinetic energy of an electron is equal to $\displaystyle\frac{{{m}_{{{e}{l}}}{{v}}^{{2}}}}{{2}},$ and it is given that the speed is the same as the minimum uncertainty of a speed. So the kinetic energy is equal to

$\displaystyle{E}_{{k}}=\frac{{{m}_{{{e}{l}}}{{v}}^{{2}}}}{{2}}=\frac{{{m}_{{{e}{l}}}{{\left(\delta{v}\right)}}^{{2}}}}{{2}}=\frac{{m}_{{{e}{l}}}}{{2}}\cdot{{\left(\frac{{h}}{{{4}\pi}}\cdot\frac{{1}}{{{m}_{{{e}{l}}}\cdot\delta{x}}}\right)}}^{{2}}=\frac{{1}}{{2}}{{\left(\frac{{h}}{{{4}\pi}}\right)}}^{{2}}\frac{{1}}{{{m}_{{{e}{l}}}\cdot{{\left(\delta{x}\right)}}^{{2}}}}.$

All values are given, and we can compute the result in Joules (note that p means pico- means $\displaystyle{{10}}^{{-{12}}}$ ):

$\displaystyle{E}_{{k}}=\frac{{1}}{{2}}\cdot\frac{{{\left({1.055}\cdot{{10}}^{{-{34}}}\right)}}^{{2}}}{{{9.11}\cdot{{10}}^{{-{31}}}\cdot{25}\cdot{{10}}^{{-{20}}}}}=\frac{{{\left({1.055}\right)}}^{{2}}}{{{2}\cdot{9.11}\cdot{25}}}\cdot{{10}}^{{-{17}}}\approx{0.00244}\cdot{{10}}^{{-{17}}}{\left({J}\right)}.$

To convert this to $\displaystyle{e}{V}$ we have to divide by $\displaystyle{e}{V}$ value:

$\displaystyle{E}_{{k}}={0.00244}\cdot\frac{{{10}}^{{-{17}}}}{{{1.60}\cdot{{10}}^{{-{19}}}}}={0.00244}\cdot\frac{{{10}}^{{2}}}{{1.60}}\approx{0.153}{\left({e}{V}\right)}.$