Question

E) Thermal broadening of emission lines

You observe four emission lines. If thermal broadening is the only source of broadening, order these lines from narrowest to broadest (take ratios):

✴ Fe II 2814 Å (T ~ 10,000 K)

✴ C II 1335 Å (T ~ 25,000 K)

✴ Si IV 1400 Å (T ~ 80,000 K)

✴ Si II 1349 Å (T~10,000 K)

Answer 1

Solution :

Doppler thermal broadening is given by :

  

For Fe II : Wavelength = 2814 Angstroms, T = 10000, m = 56 u = 56*1.6*10^(-27) kg,

k (boltzmann constant) = 1.67 * 10^(-23) J/K

   0.0268 Angstorms.

For C II :    0.043 Angstorms.

For Si IV :       0.053 Angstorms.

For Si II:    0.018 Angstorms.

Ratio : Si II < Fe ii <C II < Si IV = 1 : 1.49 : 2.39 : 2.94.