If a suitable photon is employed to locate an electron (mass \(=9.11 \times 10^{-31} \mathrm{~kg}\)) in an…
If a suitable photon is employed to locate an electron (mass \(=9.11 \times 10^{-31} \mathrm{~kg}\)) in an atom within a distance of \(10.98 \mathrm{~nm}\), the uncertainty involved in the measurement of its velocity in \(\mathrm{ms}^{-1}\) is
\(\frac{1.6565 \times 10^6}{\pi}\)
\(\frac{1.6565 \times 10^4}{\pi}\)
\(\frac{1.6565 \times 10^{-8}}{\pi}\)
\(\frac{1.6565 \times 10^8}{\pi}\)
Solution
Given,
Mass of electron \(=9.11 \times 10^{-31} \mathrm{~kg}\)
Position of electron \(=10.98 \mathrm{~nm}\)
\(=10.98 \times 10^{-9} \mathrm{~m}\)
Here, Planck constant \(=6.63 \times 10^{-34} \mathrm{Js}\)
\(\Delta v=?\)
According to Heisenberg uncertainty
\(\begin{aligned}
\Delta x. & \Delta v=\frac{h}{4 \pi m} \\
\therefore \Delta v & =\frac{6.63 \times 10^{-34} \mathrm{Js}}{4 \times \pi \times 9.11 \times 10^{-31} \mathrm{~kg} \times 10.98 \times 10^{-9} \mathrm{~m}} \\
\Delta v & =\frac{1.6565 \times 10^4}{\pi} \\
\Delta v & =\text { uncertainty in velocity of electron }
\end{aligned}\)