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

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Solution

By applying the uncertainty principle,
$Δ P = \frac{h}{4 π Δ x} = \frac{6.626 \times 10^{- 34}}{4 \times 3.1416 \times 0.002 \times 10^{- 9}} = 2.63 \times 10^{- 23} k g m / s$
The uncertainty in momentum is,
$\frac{h}{4 π m \times 0.002 \times 10^{- 9}} = \frac{h \times 5 \times 10^{11}}{4 \times π \times m}$ .

This is much larger than the actual momentum, i.e, $\frac{h}{4 π m} \times 0.05 n m$ . Hence, it cannot be defined.

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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 h4πm×0.05 nm, is there any problem in defining this value?

By applying the uncertainty principle,ΔP=h4πΔx=6.626×10−344×3.1416×0.002×10−9=2.63×10−23 kg m/sThe uncertainty in momentum is, h4πm×0.002×10−9=h×5×10114×π×m. This is much larger than the actual momentum, i.e, h4πm×0.05 nm. Hence, it cannot be defined.

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