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If uncertainties in the measurement of position and momentum of an electron are found to be equal in magnitude, what is the uncertainty in the measurement of velocity of the electron? Comment on the result obtained. |
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Answer» According to Heisenberg uncertainty principle, = \(\Delta \mathrm{x}\times\Delta p\) = \(\frac{h}{4\pi}\) Given, \(\Delta\mathrm{x}=\Delta p\) \(\therefore\) \((\Delta p)^2\) = \(\frac{h}{4\pi}\) ∆p = \(\sqrt{\frac{h}{4\pi}}\) or m x ∆v = \(\sqrt{\frac{h}{4\pi}}\) or ∆v = \(\frac{1}{m}\sqrt{\frac{h}{4\pi}}\) = \(\frac{1}{9.1\times10^{-31}}\sqrt{\frac{6.626\times10^{-34}}{4\times3.14}}\) = \(\frac{0.726\times10^{-17}}{9.1\times10^{-31}}\) = 7.98 × 1012 ms−1 Hence, uncertainty in velocity is greater than the velocity of light which is impossible. Thus, the two uncertainties cannot be equal in magnitude. |
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