University:
University of Phoenix
Course:
CHM 150 | General Chemistry I
Academic year:

2023
Views:

200

Pages:

3
Author:

Judenhutapm

Heisenberg Uncertainty Principle: It is impossible to measure simultaneously both position and momentum of a moving microscopic particle with absolute accuracy. Mathematically, The product of uncertainty in position (Δx) and uncertainty in momentum (ΔP) is a constant and is approximately equal to h/4π. Δ 𝑥 ⋅ Δ 𝑃 ≈ ℎ 4 𝜋 Δx⋅ΔP≈ 4π h Where, h = Planck's constant. Δ 𝑥 ⋅ Δ ( 𝑚 𝑣 ) ≈ ℎ 4 𝜋 Δx⋅Δ(mv)≈ 4π h Δ 𝑥 ⋅ Δ 𝑣 ≈ ℎ 4 𝜋 𝑚 Δx⋅Δv≈ 4πm h Where, Δv = Uncertainty in velocity. It explains the existence of a definite path of an electron around the nucleus i.e., orbit/shell. Q1. Calculate the uncertainty in position of a dust particle of mass 1mg, if uncertainty in velocity is 5.5 × 1 0 − 10 5.5×10 −10 m/s. Given: Δ 𝑥 = ? Δx=? 𝑚 = 1 𝑚 𝑔 = 1 0 − 6 𝑘 𝑔 m=1mg=10 −6 kg Δ 𝑣 = 5.5 × 1 0 − 10 𝑚 / 𝑠 Δv=5.5×10 −10 m/s According to Heisenberg uncertainty principle, Δ 𝑥 ⋅ Δ 𝑣 = ℎ 4 𝜋 𝑚 Δx⋅Δv= 4πm h Δ 𝑥 = ℎ 4 𝜋 𝑚 ⋅ Δ 𝑣 Δx= 4πm⋅Δv h = 6.626 × 1 0 − 34 𝑘 𝑔 ⋅ 𝑚 2 ⋅ 𝑠 − 1 4 × 3.14 × 1 0 − 6 𝑘 𝑔 × 5.5 × 1 0 − 10 𝑚 ⋅ 𝑠 − 1 = 4×3.14×10 −6 kg×5.5×10 −10 m⋅s −1 6.626×10 −34 kg⋅m 2 ⋅s −1 = 0.095 × 1 0 − 18 𝑚 =0.095×10 −18 m (Ans) Q2. Calculate the uncertainty in momentum of a particle of a microscopic particle, if the uncertainty in position is 6.626 × 1 0 − 5 6.626×10 −5 m. Given: Δ 𝑃 = ? ΔP=? Δ 𝑥 = 6.626 × 1 0 − 5 𝑚 Δx=6.626×10 −5 m According to Heisenberg uncertainty principle, Δ 𝑥 ⋅ Δ 𝑃 = ℎ 4 𝜋 Δx⋅ΔP= 4π h Δ 𝑃 = ℎ 4 𝜋 ⋅ Δ 𝑥 ΔP= 4π⋅Δx h = 6.626 × 1 0 − 34 𝑘 𝑔 ⋅ 𝑚 2 ⋅ 𝑠 − 1 4 × 3.14 × 6.626 × 1 0 − 5 𝑚 = 4×3.14×6.626×10 −5 m 6.626×10 −34 kg⋅m 2 ⋅s −1 = 0.079 × 1 0 − 30 𝑘 𝑔 ⋅ 𝑚 ⋅ 𝑠 − 1 =0.079×10 −30 kg⋅m⋅s −1 (Ans) Q3. The uncertainty in position and velocity of a particle are 1 0 − 10 𝑚 10 −10 m and 1 0 − 23 𝑚 / 𝑠 10 −23 m/s respectively. Calculate the mass of the particle in kg. Given: Δ 𝑥 = 1 0 − 10 𝑚 Δx=10 −10 m Δ 𝑣 = 1 0 − 23 𝑚 / 𝑠 Δv=10 −23 m/s 𝑚 = ? m=? According to Heisenberg uncertainty principle, Δ 𝑥 ⋅ Δ 𝑣 = ℎ 4 𝜋 𝑚 Δx⋅Δv= 4πm h 𝑚 = ℎ 4 𝜋 ⋅ Δ 𝑥 ⋅ Δ 𝑣 m= 4π⋅Δx⋅Δv h = 6.626 × 1 0 − 34 𝑘 𝑔 ⋅ 𝑚 2 ⋅ 𝑠 − 1 4 × 3.14 × 1 0 − 10 𝑚 × 1 0 − 23 𝑚 / 𝑠 = 4×3.14×10 −10 m×10 −23 m/s 6.626×10 −34 kg⋅m 2 ⋅s −1 = 0.527 × 1 0 − 4 𝑘 𝑔 =0.527×10 −4 kg = 0.0527 𝑘 𝑔 =0.0527kg (Ans) Q4. According to Heisenberg uncertainty principle, if the uncertainty in speed of an electron is 3.5 × 1 0 6 𝑚 / 𝑠 3.5×10 6 m/s, then uncertainty in its position (in mm) is at least: (a) 6 pm (b) 1.7 x 1 0 − 8 𝑚 10 −8 m (c) 17 fm (d) 2.8 x 1 0 − 10 𝑚 10 −10 m (e) 6.6 x 1 0 − 6 𝑚 10 −6 m Given: Δ 𝑣 = 3.5 × 1 0 6 𝑚 / 𝑠 Δv=3.5×10 6 m/s 𝑚 = 9.11 × 1 0 − 31 𝑘 𝑔 m=9.11×10 −31 kg (for e − − ) Δ 𝑥 = ? Δx=? According to Heisenberg uncertainty principle, Δ 𝑥 ⋅ Δ 𝑣 = ℎ 4 𝜋 𝑚 Δx⋅Δv= 4πm h Δ 𝑥 = ℎ 4 𝜋 ⋅ 𝑚 ⋅ Δ 𝑣 Δx= 4π⋅m⋅Δv h = 6.626 × 1 0 − 34 𝑘 𝑔 ⋅ 𝑚 2 ⋅ 𝑠 − 1 4 × 3.14 × 9.11 × 1 0 − 31 𝑘 𝑔 × 3.5 × 1 0 6 𝑚 / 𝑠 = 4×3.14×9.11×10 −31 kg×3.5×10 6 m/s 6.626×10 −34 kg⋅m 2 ⋅s −1 = 5.18 × 1 0 − 11 𝑚 =5.18×10 −11 m = 5.18 × 1 0 − 8 𝑚 𝑚 =5.18×10 −8 mm (Ans)