FUNDAMENTALS OF PHYSICS SIXTH EDITION - Wiley

Halliday ♦ Resnick ♦Walker

FUNDAMENTALS OF PHYSICS SIXTH EDITION Selected Solutions Chapter 40 40.11 40.23 40.27 40.49

11. The energy levels are given by En = n2 h2 /8mL2 , where h is the Planck constant, m is the mass of an electron, and L is the width of the well. The frequency of the light that will excite the electron from the state with quantum number ni to the state with quantum number nf is f = ∆E/h = (h/8mL2 )(n2f −n2i ) and the wavelength of the light is c 8mL2 c λ= = . f h(n2f − n2i ) We evaluate this expression for ni = 1 and nf = 2, 3, 4, and 5, in turn. We use h = 6.626 × 10−34 J·s, m = 9.109 × 10−31 kg, and L = 250 × 10−12 m, and obtain 6.87 × 10−8 m for nf = 2, 2.58 × 10−8 m for nf = 3, 1.37 × 10−8 m for nf = 4, and 8.59 × 10−9 m for nf = 5.

23. Schr¨ odinger’s equation for the region x > L is d2 ψ 8π 2 m + [E − U0 ] ψ = 0 . dx2 h2 If ψ = De2kx , then d2 ψ/dx2 = 4k 2 De2kx = 4k 2 ψ and 8π 2 m d2 ψ 8π 2 m 2 + [E − U ] ψ = 4k ψ + [E − U0 ] ψ . 0 dx2 h2 h2 This is zero provided

π
2m (U0 − E) . h The proposed function satisfies Schr¨ odinger’s equation provided k has this value. Since U0 is greater than E in the region x > L, the quantity under the radical is positive. This means k is real. If k is positive, however, the proposed function is physically unrealistic. It increases exponentially with x and becomes large without bound. The integral of the probability density over the entire x axis must be unity. This is impossible if ψ is the proposed function. k=

27. The energy levels are given by Enx ,ny


 
2 n2y n h2 h2 n2x y n2x + = + 2 = 8m L2x Ly 8mL2 4

where the substitutions Lx = L and Ly = 2L were made. In units of h2 /8mL2 , the energy levels are given by n2x + n2y /4. The lowest five levels are E1,1 = 1.25, E1,2 = 2.00, E1,3 = 3.25, E2,1 = 4.25, and E2,2 = E1,4 = 5.00. It is clear that there are no other possible values for the energy less than 5. The frequency of the light emitted or absorbed when the electron goes from an initial state i to a final state f is f = (Ef − Ei )/h, and in units of h/8mL2 is simply the difference in the values of n2x + n2y /4 for the two states. The possible frequencies are 0.75 (1,2 −→ 1,1), 2.00 (1,3 −→ 1,1), 3.00 (2,1 −→ 1,1), 3.75 (2,2 −→ 1,1), 1.25 (1,3 −→ 1,2), 2.25 (2,1 −→ 1,2), 3.00 (2,2 −→ 1,2), 1.00 (2,1 −→ 1,3), 1.75 (2,2 −→ 1,3), 0.75 (2,2 −→ 2,1), all in units of h/8mL2 .

49. According to Sample Problem 40-8, the probability the electron in the ground state of a hydrogen atom can be found inside a sphere of radius r is given by
 p(r) = 1 − e−2x 1 + 2x + 2x2 where x = r/a and a is the Bohr radius. We want r = a, so x = 1 and p(a) = 1 − e−2 (1 + 2 + 2) = 1 − 5e−2 = 0.323 . The probability that the electron can be found outside this sphere is 1 − 0.323 = 0.677. It can be found outside about 68% of the time.