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Q.i). State Heisenberg's uncertainty principle. Write its mathematical expression.
ii). A cricket ball weighing 100g is to be located within 0.1 Å. What is the uncertainty in velocity? (h = 6.6×10^-34 JS).

Nagaland NbseNagaland Board of School Education (Class XI) 2025Subjective· 5mImportance★★★★★
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Δx·Δp ≥ h/4π. For the 100 g cricket ball located within 0.1 Å, Δv ≈ 5.25×10^-23 m/s — vanishingly small, showing the principle is irrelevant for macroscopic objects.

i) Heisenberg's Uncertainty Principle: it is impossible to simultaneously determine, with absolute accuracy, both the exact position and the exact momentum (or velocity) of a microscopic (quantum-scale) particle such as an electron. The more precisely one quantity is known, the less precisely the other can be known. Mathematically:

Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

where Δx\Delta x = uncertainty in position and Δp\Delta p = uncertainty in momentum.

ii) Numerical: Given m=100 g=0.1 kgm = 100\,g = 0.1\,kg, Δx=0.1 A˚=0.1×10−10 m=1×10−11 m\Delta x = 0.1\,\text{Å} = 0.1 \times 10^{-10}\,m = 1\times10^{-11}\,m, h=6.6×10−34 Jsh = 6.6\times10^{-34}\,Js.

Since Δp=mΔv\Delta p = m\Delta v:

Δv≥h4π⋅Δx⋅m\Delta v \geq \frac{h}{4\pi \cdot \Delta x \cdot m}

Δv≥6.6×10−344π×(1×10−11)×0.1\Delta v \geq \frac{6.6\times10^{-34}}{4\pi \times (1\times10^{-11}) \times 0.1}

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