• 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)                 \[5.0\times {{10}^{-26}}\,kg\,m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

                    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.,        \[5\times {{10}^{-26}}\,kg\,m{{s}^{-1}}\]

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