Download F = mv r

The Bohr Atom (for hydrogen and hydrogen-like atoms)
v
Me
r
+ M
n
Positive nucleus w/ electron orbiting about in a circle (m will refer to electron mass unless
noted) Classical physics says that the electron will radiate (lose energy) and fall into the
nucleus, but it doesn't. Why this is so is answered by "old" quantum theory.
Bohr put forward four postulates of his mechanics:
(1) an e- in an atom is only allowed certain energies known as stationary states
(2) radiation is due to changes in stationary states
(3) an e- moves in circular orbits
(4) allowed states have angular momentum in multiples of nh/2π
Two important parameters may be derived given these postulates (radius, r, and energy, E).
The qualitative results of the model carry over into the modern version of q.m. so we will
examine it in detail. First, we find an expression for r
angular momentum ≡ mvr, but by postulate 1, it is given by nh/2π
∴ mvr = nh/2π
(1)
the atom contains charged particles, so Coulombic forces must be operative
Fcoul=
Z•e2
r2
(2)
where Z•e is the total positive charge and e is the fundamental charge (assuming only one
electron). Z is left in to obtain a general expression for one-electron ions such as He+ and
Li2+ where Z≠ 1.
The Coulombic force must be same as the centripetal force any body experiences in an orbit
about a central point. The centripetal force is directed toward the center and is given by
mv 2
F=
r
(3)
Equate the two forces
mv2 = Ze2 or Ze2 = mv2
r
r
r2
(4)
Use 1 to eliminate v2 and the final result is:
n2 h2
0.53n 2 Å a0 n 2
=
=
Z
Z
4 2 mZe 2
(5)
Where the Angstrom is defined as 1Å = 10-8 cm
Now look at energy:
ETotal = PE + KE
2
1
2
=-Ze
r + 2 mv
But 4 gives an expression for 1 mv2
2
2 1 Ze2
-1 Ze2
E=-Ze
r +2 r =- 2 r
(6)
(7)
and 5 gives an expression for r
2 2
The final result is E = -1 Z e
2 aon2
Notes:
(1) energy is quantized with quantum number, n, an integer =
1, 2, 3,.....
(2) as n→ ∞, Ε→ 0. The zero of energy is the ionized state (separated
electron and nucleus)
(3) Emitted radiation is given by ∆E = Einitial -Efinal
2 2
∆ E = -1 Zae ( 1 - 1 )
2 o n2f n2i
The emission lines we see in a spectrum are due to changes in energy levels
(stationary states). We can compute these wavelengths from Bohr's result for the
energy levels of an electron in an atom
2 2
E =-1 ZE
i
2 aon2i
and the spectral line is given by ∆E = Ei - Ef
2 2
∆E = 1 Zae ( 1 - 1 )
2 o n2f n2i
1 Z2e2 ≡ Rz or Rz = Z2R
where 2 ao
And R is known as the Rydberg constant and is calculated with Z = 1 (hydrogen
atom)
2
(1.60 x 10-19C)2 (9 x 109 J⋅ m)
R = 1 ea = 1 ⋅
x
2 o 2 (0.530 x 10-10m)
c2
R = 2.18 x 10-18J
or by using energy conversion factors:
R = 109678 cm-1 = 13.60 eV = 313.6 kcal mole-1
There is one more detail - we have been working with the assumption that we know
ni & n f and want to obtain ∆E or λ - we could also perform an experiment that
measures λ. Using Planck's law
∆E = hc
λ
we could obtain ∆E ( and ni, if nf is already known). All of the quantum theory we
have done is applicable. You must be able to work in all possible directions using
these expressions.