jacob_cannell comments on jacob_cannell’s Shortform

[jacob_cannell]

Velocity uncertainty of an electron

(A test of latex of sorts)

The uncertainty principle bounds the combined product of std deviations of position and momentum:

${\sigma }_x{\sigma }_p>=\frac{\hslash }{2}$

The momentum uncertainty is :

${\sigma }_p>=\frac{\hslash }{2{\sigma }_x}$

As $p=m_{e}v$, the velocity uncertainty (std deviation) is thus:

${\sigma }_v>=\frac{\hslash }{2{\sigma }_x{m}_e}$

For a 1eV electron with ${\sigma }_x\approx 1nm$ (the electron is confined to a ~1nm cavity on order of de broglie wavelength), then:

${\sigma }_v>=\frac{6.5\ast {10}^{-16}eV\ast s}{2\ast 1nm\ast 0.5MeV/c^2}$

${\sigma }_v>=\frac{6.5\ast {10}^{-16}eV\ast s}{2\ast {10}^{-9}m\ast 0.5\ast {10}^{6}eV/c^2}$

${\sigma }_v>=\frac{6.5\ast {10}^{-16}eV\ast s}{{10}^{-3}m\ast eV/c^2}$

${\sigma }_v>=\frac{6.5\ast {10}^{-16}eV\ast s}{{10}^{-3}m\ast eV/(3\ast {10}^{8}m/s{)}^2}$

${\sigma }_v>=\frac{6.5\ast {10}^{-16}eV\ast s}{{10}^{-3}m\ast eV/(9\ast {10}^{16}{m}^2/s^2)}$

${\sigma }_v>=\frac{6.5\ast {10}^{-16}eV\ast s}{1.11\ast {10}^{-20}eV\ast s^2/m}$

${\sigma }_v>=5.85\ast {10}^{4}m/s$

Meanwhile the mean or expectation of the 1eV electron’s kinetic velocity is $5.9\ast {10}^{5}m/s$ ….

So the angular std dev and or linear velocity std dev on is on order ~ 10%?