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In an atom, an electron is moving with a speed of $600 \mathrm{~m} / \mathrm{s}$ with an accuracy of $0.005 \%$. Certainty with which the position of the electron can be located is
(Given, $h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$, mass of electron $e_m=9.1 \times 10^{-31} \mathrm{~kg}$ )
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(Given, $h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$, mass of electron $e_m=9.1 \times 10^{-31} \mathrm{~kg}$ )
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The correct answer is:
$1.92 \times 10^{-3} \mathrm{~m}$
Given that velocity of electron, $v=600 \mathrm{~m} / \mathrm{s}$
Accuracy of velocity $=0.005 \%$
$\therefore \quad \Delta v=\frac{600 \times 0.005}{100}=0.03$
According to Heisenberg's uncertainty principle
$\Delta x . m \Delta v \geq \frac{h}{4 \pi}$
Or, $\begin{aligned} \Delta x & =\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.03} \\ & =1.92 \times 10^{-3} \mathrm{~m}\end{aligned}$
Accuracy of velocity $=0.005 \%$
$\therefore \quad \Delta v=\frac{600 \times 0.005}{100}=0.03$
According to Heisenberg's uncertainty principle
$\Delta x . m \Delta v \geq \frac{h}{4 \pi}$
Or, $\begin{aligned} \Delta x & =\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.03} \\ & =1.92 \times 10^{-3} \mathrm{~m}\end{aligned}$
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