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If the position of the electron is measured within an accuracy of +- 0.002nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h//(4pi xx 0.05)nm, is there any problem in defining this value? |
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Answer» Solution :`Delta x = 0.002 nm = 2 xx 10^(-3) nm = 2 xx 10^(-12) m` `Delta x xx Delta p = (h)/(4pi) " " :. Delta p = (h)/(4pi Delta x) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx (2 xx 10^(-12) m)) = 2.638 xx 10^(-23) kg m s^(-1)` Actual momentum `= (h)/(4pi xx 0.05 nm) = (h)/(4pi xx 5 xx 10^(-10)m) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx 5 xx 10^(-11) m)` it cannot be DEFINED as the actual magnitude of the momentum is smaller than the UNCERTAINTY. |
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