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Structure of Atom - Heisenberg Uncertainty Principle

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Heisenberg's Uncertainty Principle states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of a subatomic particle like an electron with absolute accuracy.

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The principle is a consequence of the dual nature of matter and radiation. It implies that the more precisely the position (Δx\Delta x) is known, the less precisely the momentum (Δp\Delta p) can be known, and vice versa.

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Mathematically, the product of the uncertainty in position and the uncertainty in momentum is always greater than or equal to a constant value: Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}.

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Since momentum p=mvp = m v, the equation can be rewritten as Δx⋅(mΔv)≥h4π\Delta x \cdot (m \Delta v) \geq \frac{h}{4\pi}, where mm is the mass of the particle and Δv\Delta v is the uncertainty in velocity.

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Significance: The uncertainty principle is significant only for microscopic objects (like electrons) and is negligible for macroscopic objects because the value of Planck's constant (h≈6.626×10−34 J sh \approx 6.626 \times 10^{-34} \text{ J s}) is extremely small.

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Impact on Atomic Model: It effectively ruled out Bohr's idea of fixed circular orbits for electrons, replacing the concept of 'trajectories' with 'probability' and 'orbitals'.

📐Formulae

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

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

Δx⋅Δv≥h4πm\Delta x \cdot \Delta v \geq \frac{h}{4\pi m}

💡Examples

Problem 1:

A microscopic particle of mass 9.1×10−31 kg9.1 \times 10^{-31} \text{ kg} has an uncertainty in its position equal to 10−10 m10^{-10} \text{ m}. Calculate the uncertainty in its velocity (Δv\Delta v). (Take h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s} and π=3.14\pi = 3.14).

Solution:

Given: m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg}, Δx=10−10 m\Delta x = 10^{-10} \text{ m}, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}. Using the formula: Δv≥h4πmΔx\Delta v \geq \frac{h}{4\pi m \Delta x}. Substituting values: Δv=6.626×10−344×3.14×9.1×10−31×10−10\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-10}}. Calculation: Δv≈5.79×105 m/s\Delta v \approx 5.79 \times 10^5 \text{ m/s}.

Explanation:

The uncertainty in velocity is quite high (5.79×105 m/s5.79 \times 10^5 \text{ m/s}), which shows that for an electron-sized particle, knowing the position accurately makes the velocity extremely uncertain.

Problem 2:

If the uncertainty in position and momentum are equal, what is the uncertainty in velocity?

Solution:

Given Δx=Δp\Delta x = \Delta p. From Heisenberg's Principle: Δx⋅Δp=h4π\Delta x \cdot \Delta p = \frac{h}{4\pi}. Substituting Δx=Δp\Delta x = \Delta p, we get (Δp)2=h4π(\Delta p)^2 = \frac{h}{4\pi}, so Δp=12hπ\Delta p = \frac{1}{2}\sqrt{\frac{h}{\pi}}. Since Δp=mΔv\Delta p = m \Delta v, then mΔv=12hπm \Delta v = \frac{1}{2}\sqrt{\frac{h}{\pi}}. Therefore, Δv=12mhπ\Delta v = \frac{1}{2m}\sqrt{\frac{h}{\pi}}.

Explanation:

This example demonstrates how to manipulate the uncertainty relation when specific conditions (like equality of uncertainties) are provided.

Heisenberg Uncertainty Principle - Revision Notes & Key Formulas | CBSE Class 11 Chemistry