One of the main quantum laws relates the uncertainties of position ∆x and momentum ∆p of quantum par — Physical Chemistry Chemistry Question
Quantum uncertainty
One of the main quantum laws relates the uncertainties of position ∆x and momentum ∆p of quantum particles. The uncertainty product cannot be less than a fixed value – a half of Planck’s constant:
where momentum is the product of mass and velocity: p = m v, the Planck’s constant is ℏ = 1.05×10–34 J s.
Without performing calculations arrange the following particles in the order of increasing minimal uncertainty of velocity, ∆vmin:
a) an electron in a H2 molecule;
b) a H atom in a H2 molecule;
c) a proton in the carbon nucleus;
d) a H2 molecule within a nanotube;
e) a O2 molecule in the room of 5 m width.
Model Answer
From uncertainty relation it follows:
Of all the particles listed above, a O2 molecule, (e), has the largest mass and ∆x and hence is characterized by smallest ∆vmin. In three other cases (b) – (d) the particles have a comparable mass – proton (b, c) and H2 molecule, therefore uncertainty of velocity may be determined by localization length ∆x. The uncertainty in position, ∆x, is the largest for nanotube (about 1 nm), smaller by an order of magnitude for H2 and is very small for the carbon nucleus, so that ∆vmin increases in the following order: (d) < (b) < (c).
Consider now localization of an electron in a H2 molecule. Electron mass is approximately 2000 smaller than that of proton, hence ∆vmin for the electron is larger than in cases (b) and (d). But the size of the carbon nucleus is by 100 thousand times (5 orders of magnitude) smaller than diameter of H2, therefore ∆vmin for the proton in the carbon nucleus is larger than that for the electron in H2.
The final sequence is as follows: (e) < (d) < (b) < (a) < (c).
For the first and the last particles from the list above calculate ∆vmin. Take the necessary reference data from handbooks or Internet.
Model Answer
For O2 molecule in a room of 5 m width we get:
∆vmin = 1.05×10–34 / (0.032 / 6.0×1023 × 2 × 5) = 2.0×10–6 m s–1 = 2.0 Å/s.
In the carbon nucleus the size of the proton localization area is equal to the nucleus diameter – about 4⋅10–15 m.
∆vmin = 1.05×10–34 / (0.001 / 6.0×1023 × 2 × 4×10–15) = 7.9×106 m s–1 ≈ 8000 km s–1.