 Bohr Model of the Atom. Experimental Observation of Hydrogen Line Emission In 1853, Anders Angstrom of Sweden first determined that a set of discrete.

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Bohr Model of the Atom

Experimental Observation of Hydrogen Line Emission In 1853, Anders Angstrom of Sweden first determined that a set of discrete frequencies were emitted by hydrogen In 1885, Johann Balmer of Switzerland analyzed hydrogen spectra and found a pattern which came to be known as the Balmer Series and founded an empirical fit.

E (eV) n  5 4 3 1 -13.6 -3.42 -1.51 -0.85 -0.54 Paschen Balmer Lymann

Rydberg-Ritz Formula Swedish physicist Johannes Rydberg and Swiss physicist Walter Ritz in the 1890’s developed a general formula to predict the emission of hydrogen 1/ = Ry[(1/n 2 )-(1/m 2 )] Ry=Rydberg Constant Ry=1.09737 x 10 7 m -1

Bohr’s attempt to explain the Balmer Series In 1913, Danish physicist Niels Bohr proposed a model that incorporated both classical and non-classical elements. 1. Atoms have a certain number of discrete energy states 2. The transition energy is related to frequency: hf = E 2 – E 1 3. The allowed states are integral multiples of Plank’s constant: L = nh, n=1,2,3,… (L is the angular momentum of electron orbits)

Orbital Mechanics Centripetal acceleration p = mv R F = Ze 2 /(4  o R 2 ) a = v 2 /R Note: Ze is the charge of the nucleus

F = ma = mv 2 /R = Ze 2 /(4  o R 2 ) Angular momentum in an orbit is L = mvR so, R = L/mv v = Ze 2 /(4  o L) since L = nh v n = Ze 2 /(4  o nh) Introduce a dimensionless constant called the fine-structure constant,   = e 2 /(4  o hc) = 1/137

So, v n = Z  c/n, n=1,2,3,… The atomic radius is found by using mvR=L = nh R n = nh/mv n = nh/m(Z  c/n)=(n 2 /Z)(h/mc  ) Define a o = h/mc  = 0.53 x 10 -10 m R n = (n 2 /Z)a o Energy: E = K + V V = -Ze 2 /(4  o r’)

E n = mv n 2 /2 -Ze 2 /(4  o R n ) E n = (m/2)(Z  c/n) 2 –(Ze 2 /(4  o ))(Zmc  /hn 2 ) E n = - (m/2)(Z  c/n) 2 This energy is negative because the electron is bound. Lowest energy state occurs when n=1 E 1 = - (m/2)(Z  c/1) 2 = -(0.51 MeV/(2x137 2 ) = -13.6 eV E n = - (13.6 eV)(Z/n) 2

Atomic Transitions in Bohr Model f n2 →n1 =[m(Z  c) 2 /2h]{(1/n 1 2 )-(1/n 2 2 )} (1/ n2 →n1 )=[m(Z  c) 2 /2hc]{(1/n 1 2 )-(1/n 2 2 )} Recall that Ry = [m(  c) 2 /2hc]=1.1x10 7 m -1 For Hydrogen Z=1 (1/ n2 →n1 )=Ry{(1/n 1 2 )-(1/n 2 2 )} (1/ n2 →n1 )= 1.1x10 7 m -1 {(1/n 1 2 )-(1/n 2 2 )}

Example 1 Free electrons are easily captured into an atomic orbit if their kinetic energy matches the orbital energy state. What kinetic energy would an electron have it matched the Bohr orbit for n=2 in hydrogen? Recall that: v n = Z  c/n, n=1,2,3,… Kinetic energy for n=2 = K n=2 =(mv 2 2 /2)=  2 mc 2 /8 K n=2 =0.51 eV/(8(137) 2 )= 3.4 eV

Example 2 The alpha-Lyman line of H is from the emission of radiation from the n=2 to n=1 level. What is the wavelength? Recall that: (1/ n2 →n1 )= 1.1x10 7 m -1 {(1/n 1 2 )-(1/n 2 2 )} (1/ n2 →n1 )= 1.1x10 7 m -1 {(1/1 2 )-(1/2 2 )}= (3/4) 1.1x10 7 m -1 (1/ n2 →n1 )= 0.825x10 7 m -1 (1/ n2 →n1 )= 1.21 x 10 -7 m = 121 nm

Example 3 In the Paschen series, the longest wavelength occurs in the transition from n=4 to n=3. What is this wavelength? Recall that: (1/ 4 →3 )= 1.1x10 7 m -1 {(1/3 2 )-(1/4 2 )} (1/ 4 →3 )= 1.1x10 7 m -1 (7/144) (1/ n2 →n1 )= 5.35x10 5 m -1 (1/ n2 →n1 )= 1.87 x 10 -6 m = 1870 nm

Hydrogen Like Ions If a heavy atom is striped of all but one electron, then it is hydrogen like. The difference is that the Centripetal force changes by the atomic number Z F = Ze 2 /(4  o R 2 ) So energy goes as E n = - (13.6 eV)(Z/n) 2 This differs from hydrogen because of the Z 2 factor. In practice, the electron mass must be replaced by the reduced electron mass Reduced mass = m/(1+(m/m nuc )) Where m nuc is the mass of the nucleus and m is the mass of the electron

The Bohr Model Works Well only with Hydrogen The use of Shr ö dinger’s equation predicts the atomic structure of complex atoms.

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