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In an atom,an electron is moving with a speed of 600m/s with an accuracy of 0.005%.Certainly with which the position of the electron can be located is ($h=6.6\times 10^{-34}kg)m^2s^{-1}$.mass of electron,$e_m=9.1\times 10^{-31}kg)$

$\begin{array}{1 1}(a)\;1.52\times 10^{-4}m&(b)\;5.10\times 10^{-3}m\\(c)\;1.92\times 10^{-3}m&(d)\;3.84\times 10^{-3}m\end{array}$

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$u=600 msec^{-1}$
$\therefore \Delta u=\large\frac{600\times 0.005}{100}$
$\Rightarrow 0.30msec^{-1}$
Now,$\Delta x.m.\Delta u=\large\frac{h}{4\pi}$
$\Delta x=\large\frac{6.6\times 10^{\large-34}}{4\times 3.14\times 9.1\times 10^{\large-31}\times 0.30}$
$\Rightarrow 1.92\times 10^{-3}m$
Hence (c) is the correct answer.
answered Jan 13, 2014 by sreemathi.v

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