Ph 312 Assignment 2 (Heisenberg Uncertainty Principle: Serway et al. p.182)

Answers (R. Egerton)

22. (delta)p (delta)x > hbar/2, (delta)v = (0.1%) v = (1E-3)v

p = mv so (delta)p = m (delta)v = (50E-3)(30) = 1.5 m/s

(delta)x > (hbar/2)/delta(p) = (1/2) (1.055E-34) / (1.5) = 3.5E-35 m

This is the minimum uncertainty in x ; the value is small (<< size of the object),

showing that quantum mechanics has little practical importance for macroscopic objects.

24. p = h/(lambda) so (delta)p = h / (delta)(lambda) = h / [(1E-6)(6000E-10)] = 1.1E-21

(delta)x > (hbar/2) / (delta)p = (1.055E-34) / (1.1E-21) = 0.96E-14 m

= minimum uncertainty in position

32. (a) v = 0.40 m/s, lambda = h/p = h/(mv) = (6.63E-34)/[(1.68E-27)(0.4)] = 9.86E-7 m

(b) D = slit separation = (1E-3) m, L = slit-detector distance = 10 m

First minimum corresponds to path difference = D sin(theta) = lambda/2

giving sin(theta) = (9.86E-7) / [(2)(1E-3)] = 4.93E-4, theta = 0.0282 deg

Distance off-axis = L tan(theta) = (10)(4.93E-4) = 4.93E-3 m = 4.93 mm

(c)We cannot identify which slit a neutron passed through without changing the neutron’s trajectory, which would destroy the diffraction pattern.

34. (a) K = (20E9) eV = 3.2E-9 J = (gamma-1)(m0)(c^2)

gamma = 1 + (3.2E-9) / [(9.11E-31)(3.0E8)^2] = 1 + 39.1E3 = 39.1E3

1- v^2/c^2 = 1/(1.53E9) = 6.55E-10, so v = c = 3.00E8 m/s to a good approximation

momentum = (gamma)(m0)(v) = (39.1E3)(9.11E-31)(3.00E8) = 1.068E-17 kg m/s

(b) lambda = h/p = (6.63E-34) / (1.068E-17) = 6.2E-17 m

This is more than a factor of 10 smaller than the diameter of a typical nucleus (1E-15 m),

as required to uncover structure within the nucleus.