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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 61 of 67
Marks: +1, -0
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\frac{h}{4 \pi_m \times 0.05 \text{ nm}} , is there any problem in defining this value.
Solution:  
According to uncertainty principle, Δx Δp = h4π\frac{h}{4\pi}
where, Δx = ± 0.002 nm = ± 0.002 × 10−9 m10^{-9} \text{ m}, Δp = ? ∴ Δp = h4πΔx\frac{h}{4\pi \Delta x} = 6.626×10−344×227×0.002×10−9\frac{6.626 \times 10^{-34}}{4 \times \frac{22}{7} \times 0.002 \times 10^{-9}} = 2.63 × 10−23 kg m/s10^{-23} \text{ kg m/s}
Actual momentum = h4πm⋅10.05\frac{h}{4\pi_m} \cdot \frac{1}{0.05} = 6.626×10−344×3.14×5×10−11\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 5 \times 10^{-11}} = 1.05 × 10−24 kg m/s10^{-24} \text{ kg m/s}
This value cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty in momentum, which is impossible.
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