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Q. If uncertainties in the measurement of position and momentum of an electron are equal, the uncertainty in the measurement of velocity is

CMC MedicalCMC Medical 2014

Solution:

Given, $ \Delta \,x=\Delta P $ or $ \Delta \,x=m\cdot \Delta v $ From Heisenbergs uncertainty principle, $ \Delta \,x\cdot m\cdot \Delta v=\frac{h}{4\pi } $ $ \Rightarrow $ $ m\cdot \Delta v\cdot m\Delta v=\frac{h}{4\pi } $ $ {{(\Delta v)}^{2}}=\frac{h}{4\pi {{m}^{2}}} $ $ \Rightarrow $ $ \Delta v=\frac{1}{2m}\sqrt{\frac{h}{\pi }} $ $ =\frac{1}{2\times 9.1\times {{10}^{-31}}}\sqrt{\frac{6.63\times {{10}^{-34}}}{3.14}} $ $ =7.98\times {{10}^{12}}m{{s}^{-1}} $ $ =8\times {{10}^{12}}m{{s}^{-1}} $