Heisenberg Uncertainty Principle PDF

Heisenberg Uncertainty Principle: It is impossible to know both the exact position and exact momentum of an object (particle) at the same time. ℏ ∆𝑥 ∆𝑝 ≥ 2 Problem 1: Uncertainty in position of electron ∆x =50 pm, find out...

Heisenberg Uncertainty Principle: It is impossible to know both the exact position and exact momentum of an object (particle) at the same time. ℏ ∆𝑥 ∆𝑝 ≥ 2 Problem 1: Uncertainty in position of electron ∆x =50 pm, find out the uncertainty in the momentum of the electron? Solution: Given that uncertainty in the position ∆x =50 pm ∆x =50 ×10−12 𝑚 Now from Heisenberg uncertainty principle ℏ ∆x ∆p≈ 2 ℏ 1.05×10−34 ∆p≈ 2∆x = 2×50×10−12 = 1.05 × 10−24 𝑘𝑔 𝑚/𝑠𝑒𝑐 Problem 2: Prove that Heisenberg uncertainty principle is dimensionally correct? Solution: Since from Heisenberg uncertainty principle we know that ℏ ∆x ∆p≥ 2………………..(1) Unit of ∆x is meter ⟹ [𝐿] Unit of ∆p is kg meter/sec ⟹ [𝑀𝐿𝑇 −1 ] So dimension of ∆x ∆p = [𝑀𝐿2 𝑇 −1 ] 𝐸 [𝑀𝐿2 𝑇 −2 ] Dimension of ℏ = 𝜔 = [𝑇 −1 ] = [𝑀𝐿2 𝑇 −1 ] Thus dimensions of equation (1) are balanced and this equation is dimensionally correct. Problem 3: If the position of an electron in a membrane is measured to an accuracy of 1.00μm, (a) What is the electron’s minimum uncertainty in momentum and velocity? (b) If the electron has this velocity, what is its kinetic energy in eV? Solution: (a) Given that uncertainty in the position ∆x =1 𝜇m=10−6 𝑚 Now from Heisenberg uncertainty principle ℏ ∆x ∆p≈ 2 ℏ 1.05×10−34 ∆p= m ∆v ≈ 2∆x = 2×10−6 = 5.25 × 10−29 𝑘𝑔 𝑚/𝑠𝑒𝑐  ℏ 1.05×10−34 ∆v ≈ 2 𝑚 ∆x = 2×9.1×10−31×10−6 = 57.6 𝑚/𝑠𝑒𝑐 (b) If electron velocity becomes v = 57.6 𝑚/𝑠𝑒𝑐 Then the kinetic energy of the electron m 𝑣2 9.1×10−31 ×(57.6)2 𝐾𝐸 = 2 = 2 = 1.51 × 10−27 𝐽𝑜𝑢𝑙𝑒𝑠 1.51×10 −27 𝐾𝐸 = 1.6×10−19 = 9.43 × 10−9 𝑒𝑉 = 9.43 𝑛𝑒𝑉 Problem 4: Suppose the velocity of an electron in an atom is known to an accuracy of 2 × 103 𝑚/𝑠𝑒𝑐. What is the electron’s minimum uncertainty in position? Solution: Given that uncertainty in the velocity of electron ∆v =2×103 𝑚/𝑠𝑒𝑐 Therefore the uncertainty in momentum of the electron ∆p= 𝑚∆v =9.1× 10−31 × 2 × 103 = 18.2 × 10−28 𝑘𝑔𝑚/𝑠𝑒𝑐 ℏ 1.05×10−34 ∆x ≈ 2 = = 0.0288 × 10−6 = 28.8 ≈ 29 𝑛𝑚 ∆p 2×18.2×10−28 Problem 5: The velocity of a proton in an accelerator is known to an accuracy of 0.250% of the speed of light. What is the smallest possible uncertainty in its position? Solution: Given that uncertainty in the velocity of electron ∆v =0.25% of the velocity of light c ∆v =0.0025 ×3×108 𝑚/𝑠𝑒𝑐 Uncertainty in momentum ∆p=m ∆v =1.67×10−27 × 0.0025 ×3×108 ∆p=0.012525×10−19 𝑘𝑔𝑚/𝑠𝑒𝑐 ℏ 1.05×10−34 Uncertainty in position ∆x ≈ 2 ∆p = 2×0.012525×10−19 = 4.19 × 10−15 𝜆 Problem 6: Show that the particle which has uncertainty in position ∆x = 2𝜋 where 𝜆 is de-Broglie wavelength. Show that velocity uncertainty is same order of magnitude as the velocity itself. Solution: Since from Heisenberg uncertainty principle we can write ℏ ∆x ∆p ≈ 2……………….. (1) 𝜆 Given that uncertainty in position ∆x = 2𝜋 ℏ ℏ ×2𝜋 ℎ Thus the uncertainty in velocity ∆𝑣 ≈ 2 𝑚 ∆𝑥 = 2𝑚𝜆 = 2𝑚𝜆………(1) h From de-Broglie wavelength relation λ= mv h v= m λ………(2) Thus from equation (1) and (2) the velocity and uncertainty in velocity, are same order of magnitude.