In an atom,an electron is moving with a speed | vedclass

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(B) Given,velocity $v = 600 \, m/s$ and percentage error $= 0.005 \%$.$\Delta v = \frac{0.005}{100} 	imes 600 = 0.03 \, m/s = 3 	imes 10^{-2} \, m/s

Vedclass · Accepted Answer

$1.92 	imes 10^{-3} \, m$

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(B) Given,velocity $v = 600 \, m/s$ and percentage error $= 0.005 \%$.$\Delta v = \frac{0.005}{100} 	imes 600 = 0.03 \, m/s = 3 	imes 10^{-2} \, m/s$.According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$,where $\Delta p = m \Delta v$.$\Delta x = \frac{h}{4 \pi m \Delta v}$.Substituting the values: $\Delta x = \frac{6.6 	imes 10^{-34}}{4 	imes 3.14 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-2}}$.$\Delta x = \frac{6.6 	imes 10^{-34}}{34.2264 	imes 10^{-33}} \approx 0.1928 	imes 10^{-1} \, m = 1.928 	imes 10^{-3} \, m$.

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

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