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# Given : The mass of electron is $9.11 \times 10^{-31}\;kg$ . Planck constant is $6.626 \times 10^{-34}\;Js$ the uncertainty involved in the measurement of velocity within a distance of $0.1\;A$ is :

$\begin{array}{1 1} (a)\;5.79 \times 10^{6}\;ms^{-1} \\(b)\; 5.79 \times 10^{7}\;ms^{-1} \\ (c)\;5.79 \times 10^{8}\;ms^{-1} \\ (d)\;5.79 \times 10^{5}\;ms^{-1} \end{array}$

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A)
Solution :
By heisenberg's uncertainty principle
$\Delta p \to$ uncertainty in momentum
$\Delta x \to$ uncertainty in position
$\Delta v \to$ uncertainty in velocity
$m \to$ mass of particle
Given,
$\Delta x= 0.1\;A=0.1 \times 10^{-10}m$
$m= 9.11 \times 10^{-31}\;kg$
h = planck constant $=6.626 \times 10^{-34}\;js$
$\pi= 3.14$
In uncertain position $\Delta v. \Delta x=\large\frac{h}{4 \pi m}$
$\Delta v \times 0.1 \times 10^{-10} =\large\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31}}$
$\Delta v=\large\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31}}$
$\qquad= 5.785 \times 10^{6}ms^{-1}$
$\qquad= 5.76 \times 10^6 ms^{-1}$