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Notely Physics 12
Class-12 Physics

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8 - N u m e r i c a l   1 3

Problem

The lifetime of an electron in an excited state is approximately 10−8 s10^{-8} \, \mathrm{s}. Calculate the uncertainty in its energy during this time.

Data

  • Uncertainty in time: Δt=10−8 s\Delta t = 10^{-8} \, \mathrm{s}
  • Planck’s constant: h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \, \mathrm{Js}

Prerequisite Concepts

  1. Heisenberg’s Uncertainty Principle:
ΔE Δt≥h4π \Delta E \, \Delta t \geq \frac{h}{4\pi}

where:

  • ΔE\Delta E is the uncertainty in energy,
  • Δt\Delta t is the uncertainty in time.
  1. Rearranging for ΔE\Delta E:
ΔE=h4πΔt \Delta E = \frac{h}{4\pi \Delta t}

Solution

Step 1: Calculate the uncertainty in energy (ΔE\Delta E)

Using the formula:

ΔE=h4πΔt\Delta E = \frac{h}{4\pi \Delta t}

Substitute the given values:

ΔE=6.626×10−34 Js4π×10−8 s\Delta E = \frac{6.626 \times 10^{-34} \, \mathrm{Js}}{4\pi \times 10^{-8} \, \mathrm{s}} ΔE=6.626×10−341.256×10−7\Delta E = \frac{6.626 \times 10^{-34}}{1.256 \times 10^{-7}} ΔE=5.27×10−27 J\Delta E = 5.27 \times 10^{-27} \, \mathrm{J}

Answer

The uncertainty in the energy of the electron is:

ΔE=5.27×10−27 J\Delta E = 5.27 \times 10^{-27} \, \mathrm{J}