91Ó°ÊÓ

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:

$∆x∆p_xâ‰\yen \frac{h}{2}$

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

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,

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

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

$$\begin{gathered} ∆x(m∆v_x)â‰\yen \frac{h}{2} \\ ∆xâ‰\yen \frac{h}{2(m∆v_x)} \end{gathered}$$

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

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

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