source : science20.com

Heisenberg Principle

“It is impossible to measure simultaneously the position and momentum or velocity of a small moving particle with absolute accuracy”.

Heisenberg  principle is represented in equation form as follows-

(ΔP)( Δx) ≥ h/4π

ΔP =mΔV

mΔV.Δx ≥ h/4π

ΔV.Δx ≥ h/4πm

Product of uncertainty in position & momentum is greater or equal to h/4π .

Solved Numerical Problems –

Question 1. The mass of an electron is 9.1×10–31 kg. Its uncertainty in velocity is 5.7×105 m/sec. Calculate uncertainty in its position?

Ans.     m= 9×10–31

ΔV= 5.7×105 m/sec                  Δx=?

h= 6.6×10–34 Joule-Sec.

Δx.Δv ≥ h/4πm

Δx ≥ h/4πmΔv

≥ 6.6×10–34/ 4× 3.14 × 9.1 × 10–31 × 5.7 × 105

≥ 0.010×10–8

Δx ≥ 1×10–10m

uncertainty in its position  ≥ 1×10–10m

Question 2.The mass of a ball is 0.15 kg & its uncertainty in position is  10–10m. What is the value of uncertainty in its velocity?

Ans.     m=0.15 kg.      h=6.6×10-34 Joule-Sec.

Δx = 10 –10 m

Δv=?

Δx.Δv ≥ h/4πm

Δv h/4πmΔx

≥ 6.6×10-34/4 × 3.14 × 0.15 × 10–10

≥ 3.50×10–24 m / sec

uncertainty in its velocity ≥ 3.50×10–24m / sec

Question 3.The mass of a bullet is 10gm & uncertainty in its velocity is 5.25×10–26 cm/sec. Calculate the uncertainty in its position?

Ans.     m=10 gm.                    h=6.6×10–27 erg-sec

Δv = 5.25×10–26 cm / sec

Δx =?

Δx.Δv ≥ h/4πm

Δx h/4πmΔv

≥ 6.6×10-27/4×3.14×10×5.25×10–26

≥ 0.10×10–2cm

≥ 1×10–3 cm

uncertainty in  position ≥ 1×10–3 cm