Problem#1
A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 nm and its momentum component along this axis with a standard deviation of Use the Heisenberg uncertainty principle to evaluate the validity of this claim.
Answer:
Heisenberg’s Uncertainty Principles tells us that 𝜟x𝜟px ≥ ℏ/2
We can treat the standard deviation as a direct measure of uncertainty.
Here 𝜟x𝜟px ≥ (1.2 x 10-10 m)(3.0 x 10-25 kg.m/s) = 3.6 x 10-35 J.s. but ℏ/2 = 5.28 x 10-35 J.s
Therefore 𝜟x𝜟px ≥ ℏ/2 so the claim is not valid.
The uncertainty product ΔxΔpx must increase by a factor of about 1.5 to become consistent
with the Heisenberg Uncertainty Principle.
Problem#2
(a) The x-coordinate of an electron is measured with an uncertainty of What is the x-component of the electron’s velocity, vx, if the minimum percentage uncertainty in a simultaneous measurement of vx is 1.0%? (b) Repeat part (a) for a proton.
Answer:
Apply the Heisenberg Uncertainty Principle, 𝜟x𝜟px ≥ ℏ/2
(a) (Δx)(mΔvx) ≥ = ℏ/2 and setting Δvx = (0.010)vx and the product of the uncertainties equal
to ℏ/2 = (for the minimum uncertainty) gives vx = ℏ/[2m(0.010)Δx] = 29.0 m/s.
(b) Repeating with the proton mass gives 15.8 mm/s.
Problem#3
An atom in a metastable state has a lifetime of 5.2 ms. What is the uncertainty in energy of the metastable state?
Answer:
Apply the Heisenberg Uncertainty Principle in the form 𝜟E𝜟t = ℏ/2
Let Δt = 5.2 x 10-3 s, the lifetime of the state of the atom, and let ΔE be the uncertainty in the
energy of the state
𝜟E > ℏ/2𝜟t = (1,055 x 10-34 J.s)/(2 x 5.2 x 10-3 s) = 1.01 x 10-32 J = 6.34 x 10-14 eV
The uncertainty in the energy is a very small fraction of the typical energy of atomic states,
which is on the order of 1 Ev.
Problem#4
(a) The uncertainty in the y-component of a proton’s position is 2.0 x 10-12 m. What is the minimum uncertainty in a simultaneous measurement of the y-component of the proton’s velocity? (b) The uncertainty in the z-component of an electron’s velocity is 0.250 m/s. What is the minimum uncertainty in a simultaneous measurement of the z-coordinate of the electron?
Answer:
The Heisenberg Uncertainty Principle says 𝜟x𝜟px ≥ ℏ/2. The minimum allowed 𝜟x𝜟px is ℏ/2. 𝜟px = m𝜟vx.
(a) 𝜟xm𝜟vx = ℏ/2
𝜟vx = ℏ/2m𝜟x = (1.055 x 10-34 J.s)/[2(1.67 x 10-27 kg)(2.0 x 10-12 m)] = 1.6 x 104 m/s
(b) 𝜟x = ℏ/2m𝜟vx = (1.055 x 10-34 J.s)/[2(9.11 x 10-31 kg)(0.250 m)] = 2.3 x 10-4 m
The smaller Δx is, the larger Δvx must be.
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