Given: The mass of electron is $9.11 \times$ $10^{-31} \mathrm{~kg}$, Planck constant is $6.626 \times$…

Given: The mass of electron is $9.11 \times$ $10^{-31} \mathrm{~kg}$, Planck constant is $6.626 \times$ $10^{-34} \mathrm{Js}$. The uncertainty involved in the measurement of velocity within a distance of $0.1 Å$ is:
  1. $5.79 \times 10^5 \mathrm{~ms}^{-1}$
  2. $5.79 \times 10^6 \mathrm{~ms}^{-1}$
  3. $5.79 \times 10^7 \mathrm{~ms}^{-1}$
  4. $5.79 \times 10^8 \mathrm{~ms}^{-1}$

Solution

$\Delta x=0.1 Å=0.1 \times 10^{-10} \mathrm{~m}$ $m=9.11 \times 10^{-31} \mathrm{~kg}$ $h=6.626 \times 10^{-34} \mathrm{Js}$ In uncertain position, $\begin{aligned} & \Delta v \times \Delta x=\frac{h}{4 \pi m} \\ & \Delta v \times 0.1 \times 10^{-10}= \\ & \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31}} \\ & \Delta v= \\ & \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 0.1 \times 10^{-10}} \\ & =5.785 \times 10^6 \mathrm{~ms}^{-1} \\ & \end{aligned}$

Asked in: NEET 2006

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