A measurement of an electron's speed is `v = 2.0 x 10^6 m/s` and has an uncertainty of `10%.` What is the minimum uncertainty in its position? (`h...

Thursday, August 7, 2014

A measurement of an electron's speed is $\displaystyle{v}={2.0}{x}{{10}}^{{6}}\frac{{m}}{{s}}$ and has an uncertainty of $\displaystyle{10}%.$ What is the minimum uncertainty in its position? ( $\displaystyle{h}\ldots$

Hello!

The uncertainty principle (one of them) states that it is impossible to exactly measure the speed and the position of a body. This inexactness is called uncertainty. Of course it is significant only for very small particles of matter, for example for electrons.

The main formula for the speed-position uncertainty is

$\displaystyle\delta{p}\cdot\delta{x}\geq\frac{{\overline{{h}}}}{{2}},$

where $\displaystyle\delta{p}$ is the uncertainty of the momentum, $\displaystyle\delta{x}$ is the uncertainty of the position and $\displaystyle{\overline{{h}}}$ is the reduced Plank's constant $\displaystyle\frac{{h}}{{{2}\pi}}.$

The momentum is the mass multiplied by the speed and $\displaystyle{10}%$ is $\displaystyle{0.1}.$ Thus we obtain the inequality

$\displaystyle{0.1}\cdot{m}_{{{e}{l}}}\cdot{v}\cdot\delta{x}\geq\frac{{h}}{{{4}\pi}},$

and therefore

$\displaystyle\delta{x}\geq\frac{{h}}{{{4}\pi}}\cdot\frac{{10}}{{{m}_{{{e}{l}}}\cdot{v}}}=\frac{{{6.626}\cdot{{10}}^{{-{34}}}}}{{{4}\pi}}\cdot\frac{{10}}{{{9.11}\cdot{{10}}^{{-{31}}}\cdot{2}\cdot{{10}}^{{6}}}}=$

$\displaystyle=\frac{{{6.626}}}{{{4}\pi\cdot{9.11}\cdot{2}}}\cdot{{10}}^{{-{8}}}\approx{0.0289}\cdot{{10}}^{{-{8}}}={2.89}\cdot{{10}}^{{-{10}}}{\left({m}\right)}.$