*1.The Rutherford Model of the Atom.*

*Fig. 1: Basic Layout of the Geiger-Marsden Experiment*Fig. 2: The Rutherford Model of the atomThe key consequence was that the
2
.. The Bohr Model of the Atom
The Bohr Model of the atom, proposed by Neils Bohr, directly challenged the
Fig. 3: The Bohr Model of the atom.From the diagram the electron (e) orbits at a radius r from the central nucleus of charge Ze. As with the planets, a centripetal (inner directed force) F acts toward the center.
m vr = nh/ 2p = n ħ
where ħ = h/ 2p is the Planck constant divided by 2p.
The Planck constant, first proposed by Max Planck, is:
h = 6.626069 x 10
^{- 34} J-s
Then the value of ħ = 1.0546 x 10
^{- 34} J-s
Next: both sides are squared:
(m vr )
^{2} = (n ħ)^{2}
So:
m^{2} v^{2 }r ^{2} = n^{2} ħ^{2}
And:
v^{2 } = n^{2} ħ^{2 }/ m^{2 }r ^{2}
Now, Bohr looked at the total energy of the H-atom in terms of it kinetic (K) and potential (V) contributions, so:
E = K + V =
½ m v^{2 } - k e^{2 / }r
E = K + V =
k e - ^{2 / }2r k e ^{2 / }r = - k e^{2 / }2r
(Since
½ m v )^{2 } = k e^{2 / }2r
Now solve for r (actually the
quantized r _{n} ):
r
_{n } = [ n^{2} ħ^{2}/ m^{2} v^{2} ] ^{½}
But from the kinetic energy equivalence:
v
^{2} = k e^{2 / }mr = n^{2} ħ^{2 }m^{2 }r ^{2}\ r _{n } = [n^{2} ħ^{2 }/ m k e^{2 } ] The Bohr radius is just the value when the principal quantum number n = 1, so :r [_{o } = ħ^{2 }/ m k e^{2 } ] = 0.0529 nm =5.2917 ×10 ^{−11} mThis is just the most probable radius, i.e. distance between proton and electron, in the hydrogen ground state.
Now, to obtain the quantized energy (E
_{n}) we substitute the value for r _{n } into the total energy equation:
E =
- k e^{2 / }2r =- k e^{2 / }2[n^{2} ħ^{2 }/ m k e^{2 }]
E = - m k
^{2 }e^{4 / }2n^{2} ħ^{2 }=- m k ^{2 }e^{4 / }2 ħ^{2 } [1/ n^{2}] = - 13.6/ n^{2 } ^{ }
Where the last quantity is in
, or eVelectron volts. Here the n refers to the energy level, ground state is n = 1, so can allow the computation of energy for a given level. Or, the energy for a photon emitted from an atom when an electron makes a transition – say from n = 2 to n = 3. Such a situation is shown below: |

**Fig. 4: A few energy (electron) transitions made in Hydrogen**

**An important point is that the**

**quantized angular momentum postulate**(m vr = n ħ) restricts the possible circular orbits to defines sizes according to the quantized radii (r

_{n }etc.). Thus the normal state of the atom, say hydrogen, will be that for which it has the least energy or the ground state – corresponding in the case of hydrogen to the Bohr radius. Some transitions for different spectral series are shown below:

*Fig. 5. Some Energy transitions in the Hydrogen Bohr atom*
As
shown in Fig. 4, emission occurs when an electron in the atom, say hydrogen,
makes a transition from a higher to a lower energy level, accompanied by the
emission of a photon with a defined energy E = hf = h (c/ l). Consider for example,
a transition from the n = 2 to the n = 1 level, as depicted in the lower right
of Fig. 4 and

*in the first line of the Lyman series*of Fig. 5.
The
energy at the n= 2 level is:

E(n=2)
= - 13.6/

**n**^{2 }**= - 13.6/**^{ }**(2)**^{2 }**= - 13.6/4 (eV)**^{ }
Now,
1 eV = 1.6 x 10

^{-19}J so:
E(n=2)
= - 13.6/4 (eV) =
-(3.4) x 1.6 x 10

^{-19}J =
-5.4 x 1.6 x 10

^{-19}J**The n= 1 level has energy:**

E(n=1)
= - 13.6/

**n**^{2 }**= - 13.6/**^{ }**(1)**^{2 }**= - 13.6 (eV)**^{ }
E(n=1)
= -(13.6) x 1.6 x 10

^{-19}J = -21.8 x 10^{-19}J**Then the energy difference is:**

E2
– E1 = [- 5.4 – (-21.8)] x 10

^{-19}J = 16.4 x 10^{-19}J
From
this, the wavelength of the photon emitted can be found. Since E = hf = h (c/ l):

l = hc/ (E2 – E1)

l = (6.626069 x 10

^{- 34}J-s)(3 x 10^{8}m/s)/ 16.4 x 10^{-19}J
l = 1.21 x 10

^{-7}m
The
frequency can be found from:

f
= (c/ l) = (3 x 10

^{8}m/s) / 1.21 x 10^{-7}m = 2.47 x 10^{15}Hz**Insight problem:**Using Fig. 5 as a basis, compute the energies and wavelengths of the photons emitted when the electron in the hydrogen atom makes the 1

^{st}, 2

^{nd}and 3

^{rd}Balmer transitions.

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