E-CHEMISTRY

 Write notes on Heisenberg's uncertainty principle. Calculate the uncertainty in velocity of a body of 1gram and also calculate the velocity of an electron.

Heisenberg's Uncertainty Principle is a fundamental concept in quantum mechanics, formulated by Werner Heisenberg in 1927. It states that:

"It is impossible to simultaneously determine both the exact position and the exact momentum of a particle with absolute certainty."

Mathematically, the principle is expressed as:

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

Where:

  • Δx\Delta x = uncertainty in position

  • Δp\Delta p = uncertainty in momentum = m⋅Δvm \cdot \Delta v

  • hh = Planck’s constant = 6.626×10−34 Js6.626 \times 10^{-34} \, \text{Js}

  • mm = mass of the particle

  • Δv\Delta v = uncertainty in velocity

So,

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

1. Calculation: Uncertainty in Velocity of a 1 gram Body

Given:

  • Mass m=1 g=1×10−3 kgm = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg}

  • Let position uncertainty Δx=1×10−9 m\Delta x = 1 \times 10^{-9} \, \text{m} (say, of atomic scale)

Δv≥h4πmΔx\Delta v \geq \frac{h}{4\pi m \Delta x}

Substitute values:

Δv≥6.626×10−344π⋅(1×10−3)⋅(1×10−9)\Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi \cdot (1 \times 10^{-3}) \cdot (1 \times 10^{-9})} Δv≥6.626×10−341.2566×10−11≈5.27×10−23 m/s\Delta v \geq \frac{6.626 \times 10^{-34}}{1.2566 \times 10^{-11}} \approx 5.27 \times 10^{-23} \, \text{m/s}

🔹 Result: The uncertainty in velocity of a 1 gram body is extremely small:
Δv≥5.27×10−23 m/s\Delta v \geq 5.27 \times 10^{-23} \, \text{m/s}

📌 This shows that for macroscopic bodies, quantum uncertainties are negligible.

2. Calculation: Velocity of an Electron

We’ll estimate the velocity of an electron by assuming it has kinetic energy equal to its typical energy in a hydrogen atom (i.e., first Bohr orbit):

E=12mv2=13.6 eV=13.6×1.602×10−19 JE = \frac{1}{2}mv^2 = 13.6 \, \text{eV} = 13.6 \times 1.602 \times 10^{-19} \, \text{J} v=2Em=2⋅13.6⋅1.602×10−199.109×10−31v = \sqrt{\frac{2E}{m}} = \sqrt{ \frac{2 \cdot 13.6 \cdot 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} } v=4.356×10−189.109×10−31≈4.782×1012≈2.19×106 m/sv = \sqrt{ \frac{4.356 \times 10^{-18}}{9.109 \times 10^{-31}} } \approx \sqrt{4.782 \times 10^{12}} \approx 2.19 \times 10^6 \, \text{m/s}

🔹 Result: The velocity of an electron (in hydrogen atom) is approximately:
v≈2.19×106 m/sv \approx 2.19 \times 10^6 \, \text{m/s}

✅ Summary

QuantityValue
Uncertainty in velocity of 1 g object (with Δx=1\Delta x = 1 nm)Δv≥5.27×10−23 m/s\Delta v \geq 5.27 \times 10^{-23} \, \text{m/s}
Velocity of an electron in hydrogen atomv≈2.19×106 m/sv \approx 2.19 \times 10^6 \, \text{m/s}

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