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

Question

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, $\frac{1.6565 \times 10^6}{\pi}$, $\frac{1.6565 \times 10^4}{\pi}$, $\frac{1.6565 \times 10^{-8}}{\pi}$, $\frac{1.6565 \times 10^8}{\pi}$

MARKS App · Accepted Answer

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 $\begin{aligned} \Delta x. & \Delta v=\frac{h}{4 \pi m} \ \therefore \Delta v & =\frac{6.63 \times 10^{-34} \mathrm{Js}}{4 \times \pi \times 9.11 \times 10^{-31} \mathrm{~kg} \times 10.98 \times 10^{-9} \mathrm{~m}} \ \Delta v & =\frac{1.6565 \times 10^4}{\pi} \ \Delta v & =\text { uncertainty in velocity of electron } \end{aligned}$

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