Question

A thin solid barrier in the $xy-$plane has a $10\mu m$diameter circular hole. An electron traveling in the $z$direction with $v_x=0m/s$passes through the hole. Afterward, is it certain that $v_x$is still zero? If not, within what range is $v_x$likely to be?

Short answer

The average velocity is zero. So, the range of velocity is from$-18m/s$to$18m/s$

Step-by-step solution

Heisenberg relation:

Heisenberg relation for position and momentum of a particle is,

$∆x∆p_x\ge \frac{h}{2}$

Rewrite the equation for $∆x$,

$∆p_x=\frac{h}{2∆x}$.....(1)

The relationship between $∆v_x$and $∆p_x$

$∆p_x=m∆v_x$.......(2)

Conversion diameter:

Convert diameter of circular hole of thin solid barrier is

$$\begin{gathered} ∆x=10\mu m \\ =10\mu m\left(\frac{{10}^{-6}m}{1\mu m}\right) \\ =10\times {10}^{-6m} \end{gathered}$$

From equations (1) and (2),

$$\begin{gathered} m∆v_x=\frac{h}{2∆x} \\ ∆v_x=\frac{h}{2m∆x} \end{gathered}$$

Substitute $6.63\times {10}^{-23}J.s$for $h,9.11\times {10}^{-31}kg$for $m$and $10\times {10}^{-6}m$for $∆x$.

$$\begin{gathered} ∆v_x=\frac{6.63\times {10}^{-23}J.s}{2\left(9.11\times {10}^{-31}kg\right)\left(10\times {10}^{-6}m\right)} \\ =36m/s \end{gathered}$$

Therefore, the average velocity is zero. So, the range of velocity is from$-18m/s$to$18m/s$