question_answer 80)

If the position of an electron is measured within an accuracy of \[\pm 0.002\,nm,\] calculate the uncertainty in momentum of the electron. Suppose the momentum of electron is \[\frac{h}{4\pi \times 0.05}\], is there any problem in defining this value?

Answer:

Given, \[\Delta x=0.002nm\] \[=2\times {{10}^{-12}}m\] \[\Delta p=\frac{h}{4\pi \Delta x}\] \[=\frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 2\times {{10}^{-12}}}\] \[=2.638\times {{10}^{-23}}kg\,m\,{{s}^{-1}}\] Momentum of electron = \[\frac{h}{4\pi \times 0.05nm}\] \[=\frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 5\times {{10}^{-11}}}\] \[=1.055\times {{10}^{-24}}kg\,m{{s}^{-1}}\] Actual value of momentum is less than the uncertainty in momentum.