Chemistry · Structure Of Atom · NEET
The formula is Δx · Δp ≥ h/4π. Here Δx is the uncertainty (error) in position and Δp is the uncertainty in momentum. Since momentum p = m·v, you can also write it as Δx · Δ(m·v) ≥ h/4π, or Δx · Δv ≥ h/(4πm) when the mass m is constant. All three forms are the same principle. NCERT lists all three as equation (2.23).
No. Δx is NOT the position itself — it is the uncertainty (the size of the error) in measuring the position. Δp is not the momentum, it is the uncertainty in momentum. So if Δx is small, you know the position very accurately. If Δx is large, the position is fuzzy. Many students wrongly plug in the actual position value instead of the uncertainty. Always read: 'uncertainty in ...' or 'error in ...'.
For NEET and NCERT chemistry, always use h/4π. So Δx · Δp ≥ h/4π. The h/2π version (called ħ, h-bar / 2, i.e. h/4π again) can appear in physics with a different constant, but the NCERT chemistry Structure of Atom formula is h/4π. Using h/2π will make your answer wrong by a factor of 2.
The product Δx · Δp can never be smaller than h/4π. It can be equal to h/4π (the best possible case, the minimum) or larger. So the real limit is the minimum value h/4π. In most numericals you use the equal sign Δx · Δp = h/4π to find the minimum uncertainty asked in the question.
Start from Δx · Δv = h/(4πm). Rearrange to Δv = h/(4π · m · Δx). Put h = 6.626×10⁻³⁴ J s, m of electron = 9.11×10⁻³¹ kg, and the given Δx in metres. The answer comes in m/s. Tip: keep everything in SI units (kg, m, s) so the units cancel cleanly.
h has units J s = kg m² s⁻¹. So Δx · Δp has units m · (kg m s⁻¹) = kg m² s⁻¹, which matches. This is why you MUST convert position to metres (not pm or nm) and mass to kg before calculating, or your units will not cancel and the answer will be off by powers of ten.
Because the right side h/4π is about 5×10⁻³⁵, an incredibly tiny number. For a heavy object like a ball, the mass m is large, so Δx · Δv = h/(4πm) becomes so small it is undetectable — you can know both position and speed. For a light electron (mass 9.11×10⁻³¹ kg), the uncertainty is large and real. That is why NEET numericals almost always use an electron or proton.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
h/4π = 6.626×10⁻³⁴ ÷ (4 × 3.14159) ≈ 5.27×10⁻³⁵ kg m² s⁻¹ (or J s). This is the minimum value of Δx·Δp.
No. Δx · Δp ≥ h/4π, and h/4π is not zero. So you can never know both position and momentum exactly at the same time. That is the whole point of the principle.
Yes. Bohr said the electron moves in a fixed circular path with a fixed radius and speed — that needs exact position and exact momentum together, which the uncertainty principle forbids. So Bohr orbits have no real meaning.
There is an energy–time form, ΔE · Δt ≥ h/4π, but for NEET Structure of Atom the main one you need is the position–momentum form Δx · Δp ≥ h/4π.
NEET regularly asks small calculations: given Δx, find Δv (or Δp), or compare uncertainty for an electron vs a heavy object. Knowing the correct formula (h/4π), the meaning of Δ, and unit conversion lets you solve these in under a minute.
