Ph 312 Assignment #1 (matter waves: Serway et al. PROBLEMS p.181)

Answers (R. Egerton)

2. (a) K = 50eV = 50(1.6E-19 J) = 80E-19 J = mv^2 / 2 classically

v^2 = 2 (80E-19) / (9.11E-31) , v = 4.19E6 m/s

lambda = h/mv = (6.63E-34) / [(9.11E-31)(4.19E6)] = 1.74E-10 m

(b) K = 50keV = (5E4)(1.6E-19 J) = 8.0E-15 J = m v^2 / 2 classically

v^2 = 2 (8.0E-15) / (9.11E-31) , v = 1.33E8 m/s

lambda = h/mv = (6.63E-34) / [(9.11E-31)(1.33E8)] = 5.47E-12 m

(using relativistic mechanics, the correct answer is lambda = 5.36E-12 m)

10. For H atom in its ground state, diameter = 2(a0) = 2 (0.53E-10 m)

= lambda = h/(mv)

v = h/[m(diameter)] = (6.63E-34) / [(9.11E-31)(1.06E-10)] = 6.86E6 m/s

K = (1/2) m v^2 = (0.5) (9.11E-31) (4.71E13) = 2.14E-17 J = 134 eV

Nearly ten times larger than the magnitude of the ground-state energy (13.6 eV)

14. (a) Constructive interference from surface atoms of spacing d requires:

n (lambda) = d sin(phi) , giving sin(phi) = n (lambda)/d = (n/d) (h/p)

Kinetic energy K = (1/2) m v^2 = [(mv)^2]/(2m) = p^2 / (2m),

where m = electron mass, so p = (2 m K)^(1/2) . Therefore,

sin(phi) = (n/d) [h/(2mK)^(1/2)] = nh/[d(2mK)^(1/2)] = nhc/[d(2m c^2 K)^(1/2)]

(b) sin(24.1deg) = 0.408, sin(54.9deg) = 0.818 . Since these values are in the ratio of 1:2 , they could correspond to n=1 and n=2 (respectively) in the diffraction formula

(in fact, these are the only two successive integers satisfying that ratio).

For K = 100eV = 1.6E-17J,

(2m c^2 K)^(1/2) = [2(9.11E-31)(3.0E8)^2 (1.6E-17)]^(1/2) = (2.62E-30)^(1/2)

= 1.62E-15 J .

Taking n=1, d = (hc)/[(0.408)(1.62E-15)] = 3.00E-10 m