The uncertainty in measuring velocity of a proton is 4×10³ m/s. The minimum uncertainty involved in measuring the position of proton will be?

Concept:

Heisenberg’s Uncertainty Principle:

It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as 

\({\rm{Δ }}x × {\rm{Δ }}{p_x} ≥ \frac{h}{{4\pi }}\)

\({\rm{Δ }}x × {\rm{Δ }}{v_x} ≥ \frac{h}{{4\pi m}}\)

where ∆x is the uncertainty in position and ∆px (or ∆vx) is the uncertainty in momentum (or velocity) of the particle.

If the position of the electron is known with high degree of accuracy (∆x is small), then the velocity of the electron will be uncertain (∆(vx) is large).

On the other hand, if the velocity of the electron is known precisely (∆(vx) is small), then the position of the electron will be uncertain (∆x will be large).

Calculation:

Given:

Δvx = 4 × 10³ m/s, h = 6.63 × 10-34 J-s;

Mass of the proton = 1.67 × 10^-27 kg;

From the uncertainty principle,

\({\rm{Δ }}x × {\rm{Δ }}{v_x} ≥ \frac{h}{{4\pi m}} ⇒ {\rm{Δ }}x ≥ \frac{h}{{4\pi m{\rm{Δ }}{v_x}}}\)

\(⇒ Δ x ≥ \frac {6.63 × 10 ^{-34}}{4\pi × 1.67 × 10^{-27}×4×10^3}\)

⇒ ΔV_x ≥  7.9 × 10^-12 m

∴ minimum uncertainity involved in measuring the position will be 7.9 × 10^-12 m.