Find the uncertainty in the position of an electron which is moving with a velocity of $2.99 \times 10^4…
Find the uncertainty in the position of an electron which is moving with a velocity of $2.99 \times 10^4 \mathrm{~cm} \mathrm{~s}^{-1}$, accurate up to $0.0016 \%$. (Given, $m_e=9.1 \times 10^{-28} \mathrm{~g}, h=6.626 \times 10^{-27}$ erg.s)
$1.211 \mathrm{~mm}$
$2.99 \times 10^{-10} \mathrm{~mm}$
$0.121 \mathrm{~mm}$
$12.11 \mathrm{~mm}$
Solution
According to Heisenberg uncertainty principle, the product of uncertainties in position $(\Delta x)$ and velocity $(\Delta v)$ is always equal to or greater than $\frac{h}{4 \pi}$.
$
\begin{aligned}
& \Delta x \cdot m \Delta v \geq \frac{h}{4 \pi} \\
& v=2.99 \times 10^4 \mathrm{cms}^{-1} \\
& \text { Uncertainty }=0.0016 \% \\
& \text { Uncertainty in velocity }(\Delta v)=2.99 \times 10^4 \times \frac{0.0016}{100} \\
& =0.478 \mathrm{~cm} \mathrm{~s}^{-1} \\
& \Delta x=\frac{h}{4 \pi m \Delta v} \\
& =\frac{6.62 \times 10^{-27} \mathrm{Js}}{4 \times 3.14 \times 9.1 \times 10^{-28} \mathrm{~g} \times 4.78 \times 10^{-1} \mathrm{~cm} \mathrm{~s}^{-1}} \\
& =12.11 \mathrm{~mm} \\
&
\end{aligned}
$