Heisenberg principle ,position &momentum 

 

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

≥6.6×10–34/9×3.14×9.1×10–31×5.7×105

≥ 0.010×10–8

≥ 1×10–10m

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

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

Δx = 10 –10 m

Δv=?

Δx.Δv ≥h/4πm

Δvh/4πmΔx

≥6.6×10-34/4×3.14×0.15×10–10

≥ 3.50×10–24m

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

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

Δv= 5.25×10–26 cm

Δx=?

Δx.Δv ≥h/4πm

Δxh/4πmΔv

≥6.6×10-34/4×3.14×10×5.25×10–26

≥ 0.10×10–2m

≥ 1×10–3 cm

Question 4.The uncertainity for the calculation of radius of the 1st Bohr Orbit is 2% for the H-atom. What will be the uncertainity in velocity of electron in the 1st Bohr Orbit (h=6.626×10–34Joule-Sec, m=9.1×10–31kg)

Ans.     r= (0.529 n*n/2 )A0

for Bohr’s 1st orbit

n=1, z=1

r    = 0.529 A

= 0.529×10–10 m