Homework 1. L φ = mωr 2 = mυr, (1)

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1 Homework 1 1. Problem: Streetman, Sixt Ed., Problem 2.2: Sow tat te tird Bor postulate, Eq. (2-5) (tat is, tat te angular momentum p θ around te polar axis is an integer multiple of te reduced Planck constant, sop θ = n ) is equivalent to an integer number of de Broglie waves fitting witin te circumference a Bor circular te orbit. Solution: (a) Te easy way: Let s write te electron angular momentum along te z axis, L φ as: L φ = mωr 2 = mυr, (1) were m is te mass of te electron, r is te radius of te orbit, is te reduced Planck constant, = /(2π), υ is te magnitude of te electron velocity, and ω = υ/r is te angular velocity. Let s write Bor s postulate as: L φ = n. (2) Ten, substituting Eq. (1) into Eq. (2): Let s multiply bot sides of tis equation by 2π/(mυ): mυr = n = n 2π. (3) 2πr = Since debroglie s assumption implies λ = /(mυ), Eq. (4) is simply wic is wat te problem asked. n mυ. (4) 2πr = nλ, (5) ECE344 Fall 29 1

2 (b) Te ard way: Let s write te tird Bor postulate as in Eq. (3): mυr = n, (6) Now, as done in class, let s consider te magnitude of te attractive electrostatic force between te electron and te nucleus and set it equal to te magnitude of te centrifugal force so tat te electron remains in a stable orbit: e 2 4πɛ r 2 = mω2 r. (7) From Eq. (7) using te definition of ω: e 2 rm 4πɛ = m 2 υ 2 r 2. (8) Ten, from Eq. (3) or (6): e 2 rm = n 2 2. (9) 4πɛ Solving for r we get discrete values labeled by te integer n: r n = 4πɛ n 2 2 me 2. (1) Now, from Eq. (6) we also get: or, using Eq. (1): mυ n = n r n, (11) mυ n = me2 4πɛ n. (12) ECE344 Fall 29 2

3 Te De Broglie wavelengt is: so tat, substituting Eq. (12) into (13): λ = mυ, (13) Comparing Eq. (1) wit Eq. (14), we see immediately tat λ = 4πɛ n me 2 = 2π n 4πɛ n 2 2 me 2. (14) 2πr n = nλ. (15) 2. Problem: Streetman (any of te two editions), Problem 2.5. For part (b) consider te form of Heisenberg uncertainty principle involving E and t, as given in te notes. Solution: (a) Since: x p = 2π, (16) for x =.1 nm we ave p = 2π x = joules s/m. (17) (b) Similarly, since: for E = 1 ev we ave: E t = 2π, (18) t = 2π E = s. (19) 3. Problem: Streetman (any of te two editions), Problem 2.6. Recall tat te resolution of any optical instrument is te ability to see small tings and it increases wit decreasing wavelengt λ. Solution: We must calculate te wavelengt of an electron wit kinetic energy of 1 ev and 12 kev and see if tis wavelengt is larger or smaller tan te wavelengt of te ligt used in an optical microscope. To do tat, ECE344 Fall 29 3

4 let s use te De Broglie assumption λ = p, (2) and calculate te electron momentum p from te fact tat E = p2 2m, (21) or: p = (2mE) 1/2, (22) so tat: λ =, (2mE) 1/2 (23) For E = 1 ev we get λ =.122 nm, wile for E =12keVwegetλ =.112 nm. Since ligt as a wavelengt in te range 4-6 nm (blue/violet to red), we see tat in eiter case te electron microscope employes muc sorter wavelengts and so it affords a muc iger resolution. 4. Problem: Streetman (any of te two editions), Problem 2.7. Solution: Te average lifetime t can be obtained as a weigted average over te distribution N(t) of te decay times t: t = tn(t)dt N(t)dt = tn e t/τ dt N e t/τ dt = t. (24) Integrating by parts te integral at te numerator: τ [(te ) t/τ ] [ τ ] t = = = τ = τ. (25) 5. Problem: Streetman, Fift Ed., Problem 2.12: Wat do Li, Na, and K ave in common? Wat do F, Cl, and Br ave in common? Wat are te electron ECE344 Fall 29 4

5 configurations for ionized Na and Cl? Te electronic sell structure of Li, Na, and K is: Li: 1s 2 2s 1. Na: 1s 2 2s 2 2p 6 3s 1. K: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1. Tey all ave one lone electron in te outermost s sell, so tat tey are very active cemically, wising to rid of tat lone electron wic is weakly bound. On te contrary, for F, Cl, and Br: F: 1s 2 2s 2 2p 5. Cl: 1s 2 2s 2 2p 6 3s 2 3p 5. Br: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4p 5. Tey all miss te last electron in te outer p sell, so tat tey are very active cemically, wising to acquire an additional electron and close te outer p sell. Tus, for Na and Cl, teir ionized form will be: Na + : 1s 2 2s 2 2p 6. Cl : 1s 2 2s 2 2p 6 3s 2 3p Problem: Streetman, Sixt Ed., Problem 2.12: Scematically sow te number of electrons in te various subsells of an atom wit te electronic sell structure 1s 2 2s 2 2p 4 and an atomic weigt of 21. Indicate ow many neutrons and ow many protons are in te nucleus. Is tis atom cemically active and wy? Solution: Since te atomic weigt A=21 is te sum of te protons and neutrons in te nucleus and te number of protons Z is equal to te number of electrons (so tat te atom remains carge-neutral overall), from te electronic configuration 1s 2 2s 2 2p 4 we get Z=8, so tat te number of neutrons will be A Z=13. Note tat tis atom is cemically active since it will be easy to accommodate 2 additional electrons in te outer p sell in order to obtain a stable octet configuration. Note: Tis atom does not exist: Te electronic configuration is tat of oxygen, but te eaviest isotope of O (radioactively unstable wit a alf-life of about 14 s) as an atomic number A=2. Yet, we know tat O is cemically very active and tis ypotetical atom would ave te same cemical properties, if it existed. ECE344 Fall 29 5