Spectroscopy

Atoms

Nucleus 99.8% of the mass \(10^{-14}m\), with total diameter \(10^{-10}m\) including electron orbits. Protons and Neutrons held together by strong nuclear force. They have mass about \(1.7*10^{-27}kg\) - about 2000 times the mass of an electron (about \(9.1*10^{-31}kg\)). The electron is held in place by electric force. Only certain orbits are allowed for electrons.

Angular momentum is quantized. $$nh = n \frac{h}{2\Pi} = mvr $$

Energy level of orbits: $$ E_n = -R_H (\frac{1}{n^2}) \tag{Eq. 5.1} $$ R is the Rydberg Constant

Rydberg Constant $$ R = 1.097*10^{7}m^{-1} \tag{Eq. 5.2} $$

Hydrogen spectrum lines between 656.3nm and 364.6nm. Spectral lines named as greeks: 1st line is \(H\alpha\), \(H\beta\), \(H\gamma\). This is the Balmer Series of lines.

Note the emission line at the red end, accounting for redness in Nebula. Balmer Series

Balmer’s formula (wavelength of spectral lines) $$ \frac{1}{\lambda} = R(\frac{1}{4} - \frac{1}{n^2}) $$

Bohr formula for Hydrogen wavelengths $$ \frac{1}{\lambda} = R(\frac{1}{N^2} - \frac{1}{n^2}) $$ (N Inner Orbit, n Outer Orbit)

N=1 is the Lyman series (ultraviolet) \(L_\alpha\)

N=3 is the Paschen series (infrared) \(P_\alpha\)

Hydrogen Spectrum Or as photo energy of energy transitions $$ hv = E_m - En = -R_h(\frac{1}{m^2} - \frac{1}{n^2}) \tag{Eq. 5.3} $$ h is planck’s constant, v here is ‘frequency’

If energy \(E > 0\) then it is a bound-free transition or free-free transition ; becomes ionised.

Excitation is absorption of a photon or collision with another atom. De-Excitation is emission of a photon.

Atoms in a given energy state or level per unit volume: $$ g_n e^{-\Epsilon/kT} $$ k is boltzmann’s constant, gn is the statistical weight

See Eq. 5.4 for more detailed info in course notes. See also Saha’s equation for the same thing with Ionisation states in a unit volume (Eq. 5.5 in the course notes)

Ionisation is where a bound-free transition occurs and electron moves to continuum state. An Ion has a nucleus with unbound electron. n =1 to E=0 is the ionisation potential of the atom.

Wavelength of \(H_\beta\)

$$ \Delta E = E_n - E_m = hv = \frac{hc}{\lambda} \tag{Eq.5.6}$$ $$ hv = E_m - En = -R_h(\frac{1}{m^2} - \frac{1}{n^2}) \tag{Eq. 5.3} $$

For a transition from 2->4 $$ \Delta E = (-0.850) - (-3.400) eV $$ $$ \lambda = \frac{hc}{\Delta E} = \frac{6.6261*10^{-34} * 2.9979*10^8}{2.550 * 1.6022 * 10^{-19}} * 10^9nm$$ $$ = 486.2nm $$