** **

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 5.7×10^{5} m/sec. Calculate uncertainty in its position?**

Ans. m= 9×10^{–31}

ΔV= 5.7×10^{5} 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}×5.7×10^{5}

≥ 0.010×10^{–8}

≥ 1×10^{–10}m

**Question 2.The mass of a ball is 0.15 kg & its uncertainty in position to 10 ^{–10}m. 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

Δv≥h/4πmΔx

≥6.6×10^{-34}/4×3.14×0.15×10^{–10}

≥ 3.50×10^{–24}m

**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

Δx≥h/4πmΔv

≥6.6×10^{-34}/4×3.14×10×5.25×10^{–26}

≥ 0.10×10^{–2}m

≥ 1×10^{–3} cm

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

Ans. r= (0.529 n*n/2 )A^{0}

for Bohr’s 1^{st} orbit

n=1, z=1

r = 0.529 A

= 0.529×10^{–10} m