Calculate the half-width of a spectral line of wavelength of λ= 550 nm when the
temperature of the gas is 6 × 105 K. Assume H atoms to be emitters.
According to http://hyperphysics.phy-astr.gsu.edu, for the Gaussian form of line
"\\dfrac{\\Delta\\lambda}{\\lambda_0} = 2\\sqrt{2\\ln2\\cdot\\dfrac{kT}{c^2}} = 2\\sqrt{2\\ln2\\cdot\\dfrac{1.38\\cdot10^{-23}\\cdot6\\cdot10^5}{(3\\cdot10^8)^2}} = 5.5\\cdot10^{-4}."
Therefore, "\\Delta\\lambda = 5.5\\cdot10^{-4}\\cdot5500\\text{\\AA} = 3\\text{\\AA} ."
Comments
Leave a comment