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If the position of the electron is measured within an accuracy of +- 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h / 4 pi m x 0.05 nm, is there any problem in defining this

2.61    If the position of the electron is measured within an accuracy of \pm 0.002\ \textup{nm}, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4\pi_m\times0.05 \ \textup{nm}, is there any problem in defining this value.

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We have given the uncertainty in position, i.e., \triangle x= \pm 0.002\ nm = 2\times10^{-12} m.

According to Heisenberg's Uncertainty Principle:

\triangle x\times \triangle p = \frac{h}{4\pi}

Where,

\triangle x is uncertainty in the position of the electron.

\triangle p is uncertainty in the momentum of the electron.

Then, \triangle p = \frac{h}{4\pi\times\triangle x}

\triangle p =\frac{6.626\times10^{-34}Js}{4\pi\times(2\times10^{-12}m)} =2.636\times10^{-23}Jsm^{-1}

Or 2.636\times10^{-23}kgms^{-1}          \left ( 1J- 1kgms^2s^{-1} \right )

The actual momentum of the electron:

\frac{h}{4\pi_{m}\times0.05\ nm}= \frac{6.626\times10^{-34}Js}{4\pi\times0.05\times10^{-9} m }

\Rightarrow p = 1.055\times10^{-24}kg\ m/s

Therefore, it cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.

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