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MCQEasyJEE 2025Heisenberg Uncertainty Principle

JEE Chemistry 2025 Question with Solution

Given below are two statements:

Statement (I): It is impossible to specify simultaneously with arbitrary precision, the linear momentum and the position of a particle.

Statement (II): If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is ≥hπ×12m\geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2m}.

In the light of the above statements, choose the correct answer from the options given below:

  • A

    Statement I is true but Statement II is false.

  • B

    Both Statement I and Statement II are true.

  • C

    Statement I is false but Statement II is true.

  • D

    Both Statement I and Statement II are false.

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two statements based on the Heisenberg Uncertainty Principle are to be checked.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

For the uncertainty relation,

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

Statement I: This is the Heisenberg Uncertainty Principle itself. It says that position and linear momentum cannot both be specified simultaneously with arbitrary precision. Hence, Statement I is true.

Statement II: If the uncertainty in position and momentum are equal, let

Δx=Δp\Delta x = \Delta p

Then from the uncertainty relation,

(Δp)2≥h4π(\Delta p)^2 \geq \frac{h}{4\pi}

So,

Δp≥h4π=12hπ\Delta p \geq \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}

For an electron,

Δp=m Δv\Delta p = m \, \Delta v

Therefore,

Δv=Δpm≥hπ×12m\Delta v = \frac{\Delta p}{m} \geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2m}

Hence, Statement II is also true.

Therefore, the correct option is B.

Checking Both Statements Separately

Given: A statement on simultaneous measurement of position and momentum, and a statement on uncertainty in velocity for an electron.

Find: Whether each statement is true or false.

  1. Use the Heisenberg Uncertainty Principle:
Δx Δp≥h4π\Delta x \, \Delta p \geq \frac{h}{4\pi}

  1. Statement I directly matches this principle, so it is true.

  2. For Statement II, assume equal uncertainties:

Δx=Δp\Delta x = \Delta p

Substituting into the uncertainty relation gives

Δx2≥h4π\Delta x^2 \geq \frac{h}{4\pi}

or equivalently,

Δp2≥h4π\Delta p^2 \geq \frac{h}{4\pi}

Thus,

Δp≥12hπ\Delta p \geq \frac{1}{2}\sqrt{\frac{h}{\pi}}
  1. Since for an electron,
Δp=m Δv\Delta p = m \, \Delta v

we get

Δv≥1m⋅12hπ=hπ×12m\Delta v \geq \frac{1}{m} \cdot \frac{1}{2}\sqrt{\frac{h}{\pi}} = \sqrt{\frac{h}{\pi}} \times \frac{1}{2m}

So Statement II is true.

Hence both statements are true, and the correct option is B.

Common mistakes

  • Confusing the Heisenberg Uncertainty Principle with limitations of instruments. The principle is fundamental, not due to experimental weakness. Treat it as an intrinsic quantum restriction on simultaneous knowledge of position and momentum.

  • Using the relation Δp=Δv\Delta p = \Delta v instead of Δp=m Δv\Delta p = m \, \Delta v. Momentum uncertainty must be converted to velocity uncertainty using the mass of the electron.

  • Missing the algebra when equal uncertainties are given. If the uncertainties are equal, substitute that condition into Δx Δp≥h4π\Delta x \, \Delta p \geq \frac{h}{4\pi} before solving for the required quantity.

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