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In an atom, an electron is moving with a speed of 800 ms-1 with an accuracy of 0.005%. Find the certainty with which the position of the electron can be located.(h = 6.626 x 10-34 Js, mass of electron = 9.1 x 10-31 kg). |
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Answer» \(\Delta v\) = 800 x \(\frac{0.005}{100}\) = 0.04 ms-1 According to Heisenberg uncertainty principle, \(\Delta \mathrm{x}\times m\Delta v\) = \(\frac{h}{4\pi}\) \(\therefore\) \(\Delta \mathrm{x}\) = \(\frac{h}{4\pi m.\Delta v}\) = \(\frac{6.626\times10^{-34}kgm^2s^{-1}}{4\times3.14\times9.1\times10^{-31}kg\times0.04ms^{-1}}\) = \(\frac{6.626\times10^{-34}}{4.572\times10^{-31}}\) = 1.449 × 103 m = 1.45 × 10−3 m |
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