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Problems · Problem 2.15

Q.A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity?

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The Heisenberg Uncertainty Principle links the uncertainty in position to the uncertainty in momentum. Given Δx=0.1 A˚\Delta x = 0.1 \, \text{Å}, the minimum uncertainty in velocity is Δv≈5.8×106 m/s\Delta v \approx 5.8 \times 10^6 \, \text{m/s}.

The core idea here is the Heisenberg Uncertainty Principle — one of the most fundamental results in quantum mechanics. It says that you cannot simultaneously know both the position and the momentum of a particle with perfect precision. The more precisely you pin down where it is, the less precisely you can know how fast it's moving (and in which direction).

In this problem, a microscope "locates" an electron within a distance of 0.1 A˚0.1 \, \text{Å}. That 0.1 A˚0.1 \, \text{Å} is the uncertainty in position, Δx\Delta x. The principle then forces a minimum uncertainty in the electron's momentum, Δp\Delta p, and from that we can find the uncertainty in velocity, Δv\Delta v.

Let's walk through it step by step.

  1. State the Uncertainty Principle The Heisenberg Uncertainty Principle for position and momentum is:

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

where hh is Planck's constant (6.626×10−34 J⋅s6.626 \times 10^{-34} \, \text{J·s}).

The ≥\ge sign means the product of the uncertainties can never be smaller than that value — it's a fundamental lower bound.

  1. Convert the given position uncertainty to metres The problem gives Δx=0.1 A˚\Delta x = 0.1 \, \text{Å}. Remember: 1 A˚=10−10 m1 \, \text{Å} = 10^{-10} \, \text{m}. So:

Δx=0.1×10−10 m=1×10−11 m.\Delta x = 0.1 \times 10^{-10} \, \text{m} = 1 \times 10^{-11} \, \text{m}.

  1. Find the minimum uncertainty in momentum To get the smallest possible Δv\Delta v, we take the equality case of the principle:

Δp=h4πΔx.\Delta p = \frac{h}{4\pi \Delta x}.

Plug in the numbers:

Δp=6.626×10−344×3.1416×1×10−11.\Delta p = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 1 \times 10^{-11}}.

First, 4π≈12.56644\pi \approx 12.5664.

Then:

Δp=6.626×10−3412.5664×10−11=6.626×10−341.25664×10−10.\Delta p = \frac{6.626 \times 10^{-34}}{12.5664 \times 10^{-11}} = \frac{6.626 \times 10^{-34}}{1.25664 \times 10^{-10}}.

Divide:

Δp≈5.27×10−24 kg⋅m/s.\Delta p \approx 5.27 \times 10^{-24} \, \text{kg·m/s}.

  1. Relate momentum uncertainty to velocity uncertainty For an electron, momentum p=mvp = m v, where mm is the electron's mass (9.11×10−31 kg9.11 \times 10^{-31} \, \text{kg}). Since the mass is known precisely, the uncertainty in momentum is directly related to the uncertainty in velocity:

Δp=m Δv⇒Δv=Δpm.\Delta p = m \, \Delta v \quad \Rightarrow \quad \Delta v = \frac{\Delta p}{m}.

Substitute:

Δv=5.27×10−249.11×10−31.\Delta v = \frac{5.27 \times 10^{-24}}{9.11 \times 10^{-31}}.

That gives:

Δv≈5.78×106 m/s.\Delta v \approx 5.78 \times 10^6 \, \text{m/s}.

  1. Interpret the result This is a huge speed — about 2% of the speed of light. It tells you that if you try to pin an electron's position down to the size of an atom (0.1 Å is roughly the diameter of a hydrogen atom), you lose almost all knowledge of its velocity. The electron could be moving anywhere from nearly stationary to millions of metres per second.
Watch out

A common mistake is to forget that Δx\Delta x must be in metres, not angstroms. Also, some students use Δp=h/Δx\Delta p = h / \Delta x (the simpler form from some textbooks), but the correct quantum mechanical lower bound is h/4πh / 4\pi. Using h/Δxh / \Delta x would give an answer about 12 times smaller — still large, but wrong for this standard formulation.

Tip

Notice that the uncertainty in velocity is enormous compared to everyday speeds. This is why we can't talk about electrons "orbiting" like planets — the uncertainty principle smears out any definite path.

✓Final answer

The minimum uncertainty in the electron's velocity is approximately 5.8×106 m/s\boxed{5.8 \times 10^6 \, \text{m/s}}.

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