1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2


 Clifton Summers
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1 Chapte 5 Example The helium atom has 2 electonic enegy levels: E 3p = 23.1 ev and E 2s = 20.6 ev whee the gound state is E = 0. If an electon makes a tansition fom 3p to 2s, what is the wavelength of the photon emitted? So λ = hc E γ = E γ = E 3p E 2s = 2.5 ev (1) 1240 ev nm 2.5 ev 500 nm (2) This is bluegeen light. Boh Model of the Hydogen Atom In the Boh model of the atom, the electons obit the nucleus like planets aound the sun. (Whethe the obits ae cicula o elliptical doesn t matte much. Boh woked out both possibilities. We will assume cicula obits fo simplicity.) Boh assumed that only some obits wee allowed. Let us conside the classical mechanics of an electon obiting a poton. We assume the poton is fixed in position and has chage +e while the electon has mass m and chage e. The foce between the poton and electon is due to Coulomb attaction: F = ke2 2 (3) whee k = 1/(4πε 0 ) = Nm 2 /C 2. The centipetal acceleation is a = v 2 /, so F = ma implies m v2 = ke2 (4) 2 o mv 2 = ke2 (5) We have 1 equation and 2 unknowns: v and. So thee is no unique value of v and, and no quantization of the enegy. The kinetic enegy is If we compae this to the potential enegy K = 1 2 mv2 = ke2 2 (6) U = ke2 whee the minus sign means that the poton and electon attact one anothe. U = 0 coesponds to =. So we see that (7) K = 1 2 U (8)
2 This is an example of the viial theoem. So the total enegy is E = K +U = 1 2 U = 1 ke 2 2 The fact that the total enegy is negative means that the electon is bound to the poton. So as the electon gets fathe and fathe away, the enegy appoaches 0. In geneal, < E < 0. To get quantized enegy levels, Boh poposed that the electon s obital angula momentum was quantized. Recall that angula momentum Fo an obiting electon, the magnitude of L is (9) L = p (10) L = mv (11) Boh poposed that the electon s obital angula momentum was quantized in intege multiples of h = h 2π = J s (12) Note that Planck s constant has units of angula momentum: and [h] = enegy time = ML2 T = ML2 T 2 T [angula momentum] = [mv] = M L ML2 L = T T Boh poposed that the electon s obital angula momentum L was quantized: L = h 2π, 2 h 2π, 3 h 2π,... = h, 2 h, 3 h,... (13) (14) = n h whee n = 1,2,3,... (15) Techically speaking, the coect theoy of quantum mechanics says that the components of L ae quantized in intege multiples of h. Fo cicula obits, L = mv, so we have L = mv = n h whee n = 1,2,3,... (16) Now we have ou second equation. So we have 2 equations (Eqs. (4) and (16)) and 2 unknowns: v and. Solving Eq. (16) fo v, we get v = n h m (17) 2
3 Plugging into Eq. (4), we find and hence, We can wite this as m whee the Boh adius a B is defined as ( ) 2 n h = ke2 m = n2 h 2 ke 2 m (18) (19) = n 2 a B whee n = 1,2,3,... (20) a B = h2 ke 2 m = nm (21) The Boh adius is half an angstom. We see that the adius of the electon obits in hydogen ae quantized as intege multiples of the Boh adius. The smallest obit (n=1) has the adius a B. Now that we know the allowed adii of the obits, we can immediately obtain the enegy associated with each obit: E = ke2 2 = 1 ke2 whee n = 1,2,3,... (22) 2a B n 2 We see that the enegies ae quantized. Each obit is associated with a paticula enegy level. Let E n denote the nth enegy level o the nth obit. When the electon makes a tansition fom n to n, a photon is emitted (if the electon deceases its enegy) o absobed (if the electon inceases its enegy). The enegy of the photon is given by ( E γ = E n E n = ke2 1 2a B n 1 ) 2 n 2 which has the same fom as the Rydbeg fomula ( 1 E γ = hcr n 1 ) 2 n 2 So Boh s model pedicts the Rydbeg fomula. Compaing these 2 fomulas gives us an expession fo the Rydbeg constant R R = ke2 2a B (hc) = (23) (24) 1.44 ev nm 2( nm)(1240 ev nm) = nm 1 (25) in pefect ageement with the obseved value. The Rydbeg o Rydbeg enegy E R is defined by E R = hcr = ke2 = m(ke2 ) 2 2a B 2 h 2 = 13.6 ev (26) 3
4 This is the binding enegy of the electon in the gound state of the hydogen atom. This is the amount of enegy you would have to give to the electon in the lowest obital in ode to libeate it fom the poton. In othe wods, this is the binding enegy. In tems of E R, the quantized enegies of the electon in the hydogen atom ae E n = E R n 2 whee n = 1,2,3,... (27) Popeties of the Boh Atom The lowest enegy state of the hydogen is called the gound state. This coesponds to n = 1 and E n=1 = E 1 = E R = 13.6 ev (28) In this state, the electon obit has the smallest adius which is the Boh adius a B : = a B = nm (29) This is the adius of a hydogen atom in its gound state and gives the ode of magnitude of the oute adius of all atoms in thei gound states. The allowed enegies o the enegy levels ae E n = E R n 2 whee n = 1,2,3,... (30) The excited states coespond to n > 1, i.e., n = 2, 3,..., e.g., E 2 = E R 4 E 3 = E R 9 = 3.4 ev = 1.5 ev An enegy level diagam looks like a ladde with the allowed enegies epesented as hoizontal lines, and the enegies inceasing as you go up. It s a good way to show tansitions between enegy levels (see Figue 5.4 in you book). So going fom n = 1 to n = 2 equies a photon with enegy equal to 10.2 ev. When electons make tansitions fom excited states (n) down to lowe enegy states (n ), a photon is emitted accoding to the fomula: ( 1 E γ = E n E n = E R n 1 ) 2 n 2 n = 2 coesponds to the Balme seies. n = 1 is called the Lyman seies, and n = 3 is the Paschen seies. The names ae fom the names of the discovees. The adius of the obit fo the nth level is given by (31) n = n 2 a B (32) Note that the adius inceases apidly with n since it goes as n 2. 4
5 We assumed cicula obits, but Boh showed that elliptical obits give the same enegy. Since this pictue of obits isn t eally coect, we won t pusue this. Example: What is the diamete of a hydogen atom with n = 100? Such atoms have been obseved in the vacuum. The diamete is d = 2 = 2n 2 a B = (0.05 nm) = 1 µm (33) HydogenLike Ions A hydogenlike ion is any atom that has lost all but one of its electons, and theefoe consists of a single electon obiting a nucleus with chage +Ze. Fo example, a He + ion o a Li 2+ ion (an electon and a lithium nucleus of chage +3e). The math fo hydogenlike ions is the same as befoe if we eplace e 2 by Ze 2. Fo example, the magnitude of the foce between the electon and the nucleus with chage Ze 2 is The potential enegy is The total enegy is F = Zke2 2 = mv2 U = Zke2 E = K +U = U 2 = Zke2 (36) 2 To find the allowed adii of an electon moving in a cicula obit aound a chage Ze, we go though the same agument as befoe: Solving fo yields L = mv = n h v = n h m mv 2 = Zke2 = m ( ) 2 n h m = n 2 h 2 Zke 2 m = n2a B Z (37) So 1 Z (38) i.e., the lage the chage Z, the smalle the adius. The bigge chage pulls the electon close. Plugging Eq. (37) into Eq. (36) gives E n = Z 2 ke2 2a B 1 n 2 (34) (35) = Z 2E R n 2 (39) 5
6 The 2 factos of Z ae easy to undestand. One comes fom the Z in the enegy and one comes fom the 1/Z in the adius. 6
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