• # question_answer                 The position of both, an electron and a helium atom is known within 1.0 mm. Further the momentum of the election is known within $\text{5}\text{.0}\times \text{1}{{\text{0}}^{-26}}\,kg\,m{{s}^{-1}}$. The minimum uncertainty in the measurement of the momentum of the helium atom is: A)                                                                                                                                                                                                 $50\,kg\,m{{s}^{-1}}$  B)                 $80\,kg\,m{{s}^{-1}}$                  C)                 $80\times {{10}^{-26}}\,kg\,m{{s}^{-1}}$            D)                 $5.0\times {{10}^{-26}}\,kg\,m{{s}^{-1}}$

By Heisenberg uncertainty principle                 $\Delta x\times \Delta p\ge \frac{h}{4\pi }$                 when the position of electron and helium atoms is known and momentum of electron is also known within a range, therefore the momentum of helium atom is equal to the momentum of electron                 i.e.,        $5\times {{10}^{-26}}\,kg\,m{{s}^{-1}}$