Heisenberg Uncertainty Principle Numericals
In Class 11 you learn that an electron is not a tiny ball going round the nucleus on a fixed path. The reason is the Heisenberg uncertainty principle. Every year, board and entrance papers ask a short numerical on it: "find the minimum uncertainty in the position of an electron" or "find the uncertainty in velocity". The calculation is only one line, but students lose marks on units and on the value of the constant. This page shows the method with the arithmetic done step by step.
The principle in simple words
You cannot know the exact position and the exact momentum of a small particle at the same time. If you pin down where the electron is (small Δx), you lose knowledge of how fast it moves (large Δp). If you know the momentum very well, the position becomes blurred. This is not a fault of the measuring instrument. It is how very small particles behave.
- Δx = uncertainty in position, in metres (m).
- Δp = uncertainty in momentum, in kg m s−1.
- Δv = uncertainty in velocity, in m s−1. For a particle of fixed mass, Δp = m Δv.
- m = mass of the particle in kg. For an electron, m = 9.109 × 10−31 kg.
- h = Planck's constant = 6.626 × 10−34 J s.
The sign ≥ means "at least". Numerical questions ask for the minimum uncertainty, so you put the equals sign and solve. Always compute the constant first:
Keep this number ready. In every problem below, the constant is the same.
Worked examples
Step 1: Δp = (h/4π) / Δx = 5.273 × 10−35 / 1.0 × 10−10 = 5.273 × 10−25 kg m s−1.
Step 2: Δv = Δp / m = 5.273 × 10−25 / 9.109 × 10−31 = 5.79 × 105 m s−1.
Step 3: Compare with the speed of light, 3.0 × 108 m s−1: 5.79 × 105 / 3.0 × 108 = 0.0019, or about 0.2 % of the speed of light.
Answer: Δv ≥ 5.8 × 105 m s−1. An electron held inside an atom must have a very large uncertainty in its velocity. So a fixed orbit cannot be defined.
Step 1: Δv = (h/4π) / (m Δx) = 5.273 × 10−35 / (0.10 × 1.0 × 10−3).
Step 2: The denominator is 1.0 × 10−4, so Δv = 5.273 × 10−31 m s−1.
Answer: Δv ≥ 5.3 × 10−31 m s−1. This is far too small to measure. So for everyday objects the uncertainty principle has no practical effect, and we can talk about path and speed together.
Step 1: Δv = 0.1 % of 3.0 × 106 = 3.0 × 106 × 0.001 = 3.0 × 103 m s−1.
Step 2: Δx = (h/4π) / (m Δv) = 5.273 × 10−35 / (9.109 × 10−31 × 3.0 × 103).
Step 3: Denominator = 9.109 × 10−31 × 3.0 × 103 = 2.733 × 10−27.
Step 4: Δx = 5.273 × 10−35 / 2.733 × 10−27 = 1.93 × 10−8 m.
Answer: Δx ≥ 1.9 × 10−8 m, which is 19 nm. That is about 190 times the 1.0 × 10−10 m size of an atom. So a fast electron whose speed is known this well cannot be said to be "inside" one atom.
Step 1: Δp = 5.273 × 10−35 / 5.0 × 10−11 = 1.05 × 10−24 kg m s−1.
Step 2: Check: halving Δx should double Δp. Example 1 had Δx = 1.0 × 10−10 and Δp = 5.27 × 10−25. Here Δx is half, and Δp = 1.05 × 10−24 is double. ✓
Answer: Δp ≥ 1.05 × 10−24 kg m s−1.
Why the electron and the ball behave so differently
The constant h/4π is extremely small, 5.27 × 10−35 J s. For a heavy object, the mass in the denominator makes the uncertainty even smaller (Example 2). For an electron, the mass is only 9.109 × 10−31 kg, so the same constant gives a very large uncertainty (Example 1). The principle is always true, but it matters only for particles as light as electrons.
What it means for the atom
Bohr's model gave the electron a fixed orbit with an exact radius and an exact speed. The uncertainty principle says this is not possible. So the modern picture replaces the orbit with an orbital: a region of space around the nucleus where the probability of finding the electron is high. This is why your Class 11 syllabus moves from the Bohr model to the quantum mechanical model.
- Using grams. The mass of the electron must be in kg: 9.109 × 10−31 kg. Using 9.109 × 10−28 g with h in J s gives a wrong answer by a factor of 1000.
- Forgetting the 4π. Writing h in place of h/4π makes the answer 12.57 times too large. Check which form your textbook and question use.
- Mixing Δp and Δv. If the question gives velocity, divide by m once. Do not put Δv directly into the Δx Δp formula.
- Using the radius as Δx. Use the number the question gives as the uncertainty in position. Do not change it to a diameter or a radius unless asked.
- Powers of ten. Write the constant and the data in standard form before you divide, and subtract the exponents carefully.
- Saying "the electron has no position". The principle says the position and momentum cannot both be known exactly together. It does not say the electron does not exist.
Textbook differences to watch
Most Class 11 books write Δx · Δp ≥ h/4π. A few older books use Δx · Δp ≈ h/2π or even h. If your teacher or book uses another form, use that form in the exam, because the numerical answer changes by a constant factor. The method, the units and the idea stay the same.
Exam relevance
| Question type | What to do |
|---|---|
| Find Δv from Δx (electron) | Δv = (h/4π) / (m Δx); m in kg |
| Find Δx from Δv or from a percentage of the speed | Find Δv first, then Δx = (h/4π) / (m Δv) |
| Why is the principle ignored for a ball? | The mass is large, so the uncertainty is far too small to notice |
| Why is the Bohr orbit not accepted? | An exact path needs exact position and momentum together |
Check your own board's syllabus and latest sample paper to see how this topic is asked. Do not assume a fixed number of marks.
Use the suite to do the powers of ten and the divisions in these numericals, then compare with your own working.
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