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Structure of Atom - Quantum Mechanical Model of Atom

Grade 11CBSEChemistry

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

🔑Concepts

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Dual Nature of Matter: Proposed by de Broglie, matter exhibits both particle and wave-like properties. The wavelength is inversely proportional to momentum: λ=hmv\lambda = \frac{h}{mv}.

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Heisenberg's Uncertainty Principle: It is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron with absolute accuracy. This is represented as Δx⋅Δp≥h4π\Delta x \cdot \Delta p \ge \frac{h}{4\pi}.

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Schrödinger Wave Equation: The fundamental equation of quantum mechanics, H^ψ=Eψ\hat{H}\psi = E\psi, where ψ\psi is the wave function representing the amplitude of the electron wave.

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Probability Density: While ψ\psi has no physical meaning, ψ2\psi^2 represents the probability density, which is the probability of finding an electron at a point within an atom.

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Quantum Numbers: Four numbers used to describe an electron's state: Principal (nn) for size/energy, Azimuthal (ll) for shape, Magnetic (mlm_l) for orientation, and Spin (msm_s) for rotation.

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Orbitals and Nodes: Orbitals are regions with high electron probability. Nodes are regions where the probability density ψ2\psi^2 is zero. Total nodes = n−1n - 1; Radial nodes = n−l−1n - l - 1; Angular nodes = ll.

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Electronic Configuration Rules: The Aufbau Principle (fill lower energy orbitals first), Pauli Exclusion Principle (no two electrons can have the same four quantum numbers), and Hund's Rule (maximize total spin multiplicity by filling degenerate orbitals singly first).

📐Formulae

λ=hmv=hp\lambda = \frac{h}{mv} = \frac{h}{p}

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

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

Orbital Angular Momentum=l(l+1)h2π\text{Orbital Angular Momentum} = \sqrt{l(l+1)} \frac{h}{2\pi}

Total Nodes=n−1\text{Total Nodes} = n - 1

Radial Nodes=n−l−1\text{Radial Nodes} = n - l - 1

Angular Nodes=l\text{Angular Nodes} = l

Max electrons in a shell=2n2\text{Max electrons in a shell} = 2n^2

💡Examples

Problem 1:

Calculate the de Broglie wavelength of an electron (mass =9.1×10−31= 9.1 \times 10^{-31} kg) moving with a velocity of 2.05×1072.05 \times 10^7 m/s.

Solution:

Given: m=9.1×10−31m = 9.1 \times 10^{-31} kg, v=2.05×107v = 2.05 \times 10^7 m/s, h=6.626×10−34h = 6.626 \times 10^{-34} Js. Using λ=hmv\lambda = \frac{h}{mv}: λ=6.626×10−349.1×10−31×2.05×107≈3.55×10−11\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.05 \times 10^7} \approx 3.55 \times 10^{-11} m.

Explanation:

The de Broglie equation relates the wave character (λ\lambda) to the particle character (mass and velocity) of matter.

Problem 2:

Determine the number of radial and angular nodes for a 4p4p orbital.

Solution:

For 4p4p: n=4n = 4 and l=1l = 1. Angular nodes =l=1= l = 1. Radial nodes =n−l−1=4−1−1=2= n - l - 1 = 4 - 1 - 1 = 2.

Explanation:

Angular nodes are determined solely by the azimuthal quantum number ll, while radial nodes depend on both nn and ll.

Problem 3:

A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.10.1 Å. What is the uncertainty in the measurement of its velocity?

Solution:

Δx=0.1\Delta x = 0.1 Å =10−11= 10^{-11} m. From Heisenberg's Principle: Δv=h4πmΔx=6.626×10−344×3.14×9.1×10−31×10−11≈5.79×106\Delta v = \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-11}} \approx 5.79 \times 10^6 m/s.

Explanation:

Because the uncertainty in position is very small (atomic scale), the uncertainty in velocity becomes very large, illustrating the limit of simultaneous measurements in quantum mechanics.

Quantum Mechanical Model of Atom - Revision Notes & Key Formulas | CBSE Class 11 Chemistry