Question

What is the minimum uncertainty in position, in $nm$, of an electron whose velocity is known to be between $3.48\times {10}^{5}m/s$and $3.58\times {10}^{5}m/s$?

Short answer

Round off to two significant figures, the minimum uncertainty in position of the electron is$36nm$

Step-by-step solution

Given Information

Calculate the minimum uncertainty in position of the electron using the Heisenberg's uncertainty principle.

According to the Heisenberg's uncertainty principle, the uncertainty in position $(∆x)$and the momentum $(∆p_x)$of the electron is as follows:

$∆x∆p_x\ge \frac{h}{2}$

Here, $h$is the Planck's constant.

Expression

The uncertainty in momentum of the electron is,

$∆p_x=m∆v_x$

Here, $m$is the mass of the electron and $∆v_x$is the uncertainty in the velocity of the electron.

The uncertainty in the velocity of the electron is,

$$\begin{gathered} ∆v_x=3.58\times {10}^{5}m/s-3.48\times {10}^{5}m/s \\ =0.10\times {10}^{5}m/s \end{gathered}$$

Substitute $m∆v_x$for $∆p_x$in the equation $∆x∆p_x\ge \frac{h}{2}$and solve for $∆x$.

$$\begin{gathered} ∆x(m∆v_x)\ge \frac{h}{2} \\ ∆x\ge \frac{h}{2(m∆v_x)} \end{gathered}$$

Substitute $6.63\times {10}^{-34}J.s$for $h.9.11\times {10}^{-31}kg$for $m$, and $0.10\times {10}^{5}m/s$for $∆v_x$.

$$\begin{gathered} ∆x\ge \frac{6.63\times {10}^{-34}J.s}{2(9.11\times {10}^{-31}kg)(0.10\times {10}^{5}m/s)} \\ =3.64\times {10}^{-8}m \\ =3.64\times {10}^{-8}m\left(\frac{1nm}{{10}^{-9}m}\right) \\ =36.4nm \end{gathered}$$

Round off to two significant figures, the minimum uncertainty in position of the electron is$36nm$