In an atom, an electron is moving with a speed of $600 m / s$ with an accuracy of 0.005 %. Certainity with which the position of the electron can be located is

( $h = 6.6 \times 10^{- 34} k g m^{2} s^{- 1}$ ,mass of electron, $e^{m} = 9.1 \times 10^{- 31} k g$ )

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Solution

The correct option is B

According to Heisenberg uncertainty principle.

$Δ x . m Δ v = \frac{h}{4 π}, Δ x = \frac{h}{4 π m Δ v}$

Here $Δ v$ = $\frac{600 \times 0.005}{100} = 0.03$

So, $Δ x = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 9.1 \times 10^{- 31} \times 0.03}$

= $1.92 \times 10^{- 3} m e t e r$

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In an atom, an electron is moving with a speed of 600 m/sec with an accuracy of 0.005%. Certainty with which the position of the electron can be located is:

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In an atom, an electron is moving with a speed of 600 $s e c^{- 1}$ with a accuracy of 0.005%, certainity with which the position of electron can be located is