The uncertainty in the position of an electron moving with velocity of \(3 \times 10^4 \mathrm{~cm} /…
The uncertainty in the position of an electron moving with velocity of \(3 \times 10^4 \mathrm{~cm} / \mathrm{s}\) is
(given mass of electron \(=9.1 \times 10^{-28}\),
uncertainty in velocity \(=\mathbf{0. 0 2} \%\))
\(1.8 \times 10^{-3} \mathrm{~cm}\)
\(9 \times 10^{-3} \mathrm{~cm}\)
\(3.8 \times 10^{-2} \mathrm{~cm}\)
\(1.8 \times 10^{-4} \mathrm{~cm}\)
Solution
Heisenberg's uncertainty equation,
\(\begin{array}{ll}
\Delta x \times \Delta P \geq \frac{h}{4 \pi} \\
\Rightarrow \quad \Delta x \times m \Delta v \geq \frac{h}{4 \pi} \\
\therefore \Delta x=\text { uncertainty in position } \\
\Delta P=m \Delta v=\text { uncertainty in momentum } \\
\Rightarrow \Delta x=\frac{h}{4 \pi \times m \times \Delta v}
\end{array}\)
\(\begin{aligned}
& =\frac{6.626 \times 10^{-27} \mathrm{erg} \mathrm{s}}{4 \times 3.14 \times\left(9.1 \times 10^{-28} \mathrm{~g}\right)} \times\left(3 \times 10^4 \times \frac{0.02}{100} \mathrm{~cm} \mathrm{~s}^{-1}\right) \\
& =9.66 \times 10^{-3} \mathrm{~cm}
\end{aligned}\)
So, in option (b) answer should be written as \(9.66 \times 10^{-3} \mathrm{~cm}\) or \(9 \times 10^{-3} \mathrm{~cm}\).