BOHR - RUTHERFORD HYDROGEN ATOM

This model of 1913 improves Rutherford's original nuclear model by placing restrictions on the possible "orbits" of the electron. It is the "Grade 9 - 10" atom that is divied up as a standard representation.

THE FOLLOWING IS NOT ON TCE SYLLABUS - notice will be given when to wake up!

We start by calculating the total energy = sum of potential and kinetic energies of the orbiting electron

and we equate Coulomb's Law with the necessary centripetal force for orbiting.

(1) F = k e² / r² = mv² / r e = charge on both electron and proton = 1.6 x 10^-19 C

From the above E_k = 1/2 mv² = k e² /2 r (2)

E_p = Work Done in moving charge from infinity to orbital radius =

= -k e² / r

so the total energy E = -k e² / 2r . (3) (Notice the energy is "negative' - that simply means that infinitely far away is taken as a "sea level" or zero.)

This equation allows any old radius and energy. Rutherford put it up. It does not solve emission lines. It also has a weakness, the electrons are accelerating ( circular motion ) thus should be radiating radio waves! - losing energy - spiralling in - exit atom.

Enter Niels Bohr

LET (for the hell of it) linear momentum x radius mvr = nh /2π where n = 1,2,3,4,....... (positive integers), h = Planck's Constant

Now to fiddle to sus out the consequences - basically we are going to return to total energy again after wiping out speed.

Square the lot, and reorganise mv² = ( nh /2π )² / mr² = k e² / r from (1)

OUT OF THIS COMES A RESTRICTION ON ORBITAL RADII

r = n²h² / 4π²kme^{2 n = 1,2,3,4,.............}

Stick this in a calculator

for n=1 r = 0.53 x 10^-11 m

n = 2 4 x this value,

n = 3 9 x this value .......

When we bung this back in the Energy equation (3) to get

E = - 2π²k²me⁴ / n²h^{2 (4)}

Energies are restricted!

Next guess by Bohr - ASSUME DOWNWARD CHANGES IN ORBITS by electrons are accompanied by light emission = energy difference between orbits

Then

ΔE = 2π²k²me⁴ (1/m² - 1/n²) = hf = hc/λ where m = 1,2,3,..... & n = 2,3,4,5,.....
h²

thus 2π²k²me⁴ (1/m² - 1/n²) = 1/λ
h³c

OH WOW! the Rydberg formula !!!! and it works really well for the spectra of hydrogen.

How to interpret this?

"Promotion" of an electron outwards requires energy. This can be due either to

• a photon exactly matching the energy necessary - in which case we have an absorption spectrum - these colours are removed to promote the electron
• an energetic electron bashing into it in a flame or from a high energy current

"Demotion" of an electron inwards releases photons exactly matching the energy changes from orbit to orbit

• Emission spectrum

Only certain orbits exist corresponding to diameters of 0.106 nm, 0.424nm, 0.954nm, ........

and the energies corresponding to these orbits are, on substitution in (4) and converting to eV,

-13.6eV, -3.4eV, -1.51eV .......

Jumps downwards between orbits lead to the spectral patterns -

• down to n = 1, energies of 13.6 - 3.4 = 10.2 eV, 13.6 - 1.51 = 11.09 eV, ........ these changes in energy correspond to photons of these energies - the series in the ultra-violet.
• down to n = 2, energies of 3.4 - 1.51 = 1.89 eV, ....... the visible series.