Question

According to the Heisenberg uncertainty principle, what is the minimum possible 1D box size (L) capable
of trapping an electron (me = 9.109 × 10−31 kg)

Answer 1

According to the Heisenberg uncertainty principle::

$\Delta$x.$\Delta$p = $\frac{h}{4\pi }$ ......................(1)

where h=6.6*10^-34 j.s

  $\Delta x$=1mm

Me = 9.1*10^-31 kg and

find   $\large \Delta v$=??

$\large \Delta$p =M_e*$\large \Delta v$.............(2)

we can calculate that $\large \Delta x$ *$\large \Delta p$ = $\large \frac{h}{4\pi }$   $\large \approx$10^-34

put the value in equation number (2)

(10^-3m)(10^-30)$\large \Delta v$ =10^-34

10^-33m.kg($\large \Delta v$)=10^-34

$\large \Delta v$=10^-1=0.1m/s

$\large \Delta v$=0.1 m/s