Heisenberg Uncertainty

Answer: B. The principle states that position and momentum cannot BOTH be simultaneously known with arbitrary precision for a microscopic particle (NCERT Class 11 Chemistry Chapter 2, page 52). It is about the simultaneous determination of the pair, not about each quantity individually.

Why A is wrong: A addresses only position; the principle concerns the SIMULTANEOUS measurement of both position and momentum, not the measurability of either alone.

Why C is wrong: C is incomplete — the principle does not say momentum is 'always uncertain' in isolation; it says the PRODUCT of position and momentum uncertainties has a minimum bound.

Why D is wrong: D refers to the energy-time uncertainty relation, which is a separate formulation and not the position-momentum statement asked here.

Answer: A. The Heisenberg uncertainty principle states Δx·Δp ≥ h/(4π). The minimum product (equality case) is h/(4π) (NCERT Class 11 Chemistry Chapter 2, page 52).

Why B is wrong: B uses h/(2π) = ℏ (reduced Planck constant); while related, the correct lower bound is h/(4π) = ℏ/2, not ℏ.

Why C is wrong: C uses h without any denominator — this is the common trap of forgetting the 4π factor entirely, giving a value ~12.6× too large.

Why D is wrong: D uses h/(8π); there is no factor of 8π in the standard Heisenberg relation — this results from incorrectly doubling the denominator.

Answer: A. A is correct. Δp = m·Δv = 1.0 × 10⁻²⁸ × 3.0 × 10⁵ = 3.0 × 10⁻²³ kg·m/s. Δx_min = h/(4π·Δp) = 6.6 × 10⁻³⁴ / (4 × 3.14 × 3.0 × 10⁻²³) = 6.6 × 10⁻³⁴ / (3.77 × 10⁻²²) = 1.75 × 10⁻¹² m ≈ 1.76 × 10⁻¹² m. Check the exponent on its own: −34 − (−22) = −12.

Why B is wrong: B is wrong because 1.76 × 10⁻¹⁰ m is a hundred times too large. It needs the denominator to come out near 10⁻²⁴, which would mean losing two powers of ten from Δp = 3.0 × 10⁻²³.

Why C is wrong: C is wrong because 1.76 × 10⁻¹¹ m is ten times too large, and it is the easiest error to make here: the division 6.6 × 10⁻³⁴ / 3.77 × 10⁻²² is carried out correctly and the exponent is then written one place off. Subtract the exponents separately as a check — −34 − (−22) = −12, not −11.

Why D is wrong: D is wrong because 1.76 × 10⁻⁹ m is a thousand times too large. Even dropping the 4π entirely gives h/(m·Δv) = 6.6 × 10⁻³⁴ / 3.0 × 10⁻²³ = 2.2 × 10⁻¹¹ m, so no single omission in the formula reaches 10⁻⁹.

Answer: A. Δv_min = h/(4π·m·Δx) = 6.6 × 10⁻³⁴ / (4 × 3.14 × 9.1 × 10⁻³¹ × 1.0 × 10⁻¹⁰) = 6.6 × 10⁻³⁴ / (1.143 × 10⁻³⁹) ≈ 5.77 × 10⁵ m/s ≈ 5.8 × 10⁵ m/s.

Why B is wrong: B (10⁶) results from converting 0.1 nm incorrectly as 10⁻¹¹ m instead of 10⁻¹⁰ m — a unit-conversion error on the nano prefix.

Why C is wrong: C (10⁷) compounds the above error with forgetting that 0.1 nm = 1 × 10⁻¹⁰ m, using 10⁻¹² m instead.

Why D is wrong: D leaves out the 4π entirely: Δv = h/(m·Δx) = 6.6 × 10⁻³⁴/(9.1 × 10⁻³¹ × 1.0 × 10⁻¹⁰) ≈ 7.3 × 10⁶ m/s.

Answer: D. The standard form is Δx·Δp ≥ h/(4π), where Δp is the momentum uncertainty (NCERT Class 11 Chemistry Chapter 2, page 52). Options using Δv instead of Δp, or using h or h/(2π) as the bound, are incorrect formulations.

Why A is wrong: A omits the 4π denominator — this is the most common recall error and gives a bound ~12.6× too large.

Why B is wrong: B uses h/(2π) = ℏ; the correct bound is h/(4π) = ℏ/2.

Why C is wrong: C replaces Δp with Δv; the principle is stated in terms of momentum uncertainty, not velocity uncertainty. While Δp = mΔv can be substituted, the standard form uses Δp.

Answer: B. Bohr's model assumes electrons travel in fixed circular orbits with well-defined radius (position) and velocity (momentum) simultaneously. The Heisenberg principle forbids such simultaneous precision — making the concept of a definite orbit invalid (NCERT Class 11 Chemistry Chapter 2, page 52).

Why A is wrong: A is true (spin is unaccounted) but is a limitation from quantum mechanics in general, not specifically from the Heisenberg uncertainty principle.

Why C is wrong: C is a real limitation of Bohr's model but stems from the inability to handle electron-electron repulsion, not from the uncertainty principle.

Why D is wrong: D is factually wrong — Bohr's model gives correct energy values for hydrogen; its failure is conceptual (definite orbits), not computational for one-electron systems.

Answer: D. Δx_min = h/(4π·Δp). Since Δp is the same for both particles and the formula depends only on Δp (not on mass), both have the same Δx_min. Mass enters only if the problem specifies Δv; when Δp itself is given as equal, mass is irrelevant.

Why A is wrong: A assumes the heavier particle has more uncertainty — but when Δp (not Δv) is given as equal, mass does not appear in Δx = h/(4π·Δp).

Why B is wrong: B assumes the lighter particle has more uncertainty in position — this would be true if Δv were equal (since Δp = m·Δv, lighter particle → smaller Δp → larger Δx), but the stem specifies equal Δp.

Why C is wrong: C is wrong because the formula Δx_min = h/(4π·Δp) directly shows that equal Δp gives equal Δx_min regardless of other quantities.

Answer: C. Δp_min = h/(4π·Δx) = 6.6 × 10⁻³⁴ / (4 × 3.14 × 2.0 × 10⁻¹⁰) = 6.6 × 10⁻³⁴ / (2.51 × 10⁻⁹) ≈ 2.63 × 10⁻²⁵ kg·m/s ≈ 2.64 × 10⁻²⁵ kg·m/s.

Why A is wrong: A uses h/(2π·Δx) instead of h/(4π·Δx) — using ℏ instead of ℏ/2 as the bound, doubling the correct answer.

Why B is wrong: B (10⁻²⁴) results from using Δx = 2.0 × 10⁻¹¹ m (misreading the exponent or converting Å to m incorrectly).

Why D is wrong: D uses h/Δx without the 4π factor — the full trap of Δp = h/Δx instead of h/(4π·Δx), yielding a value ~12.6× too large.