y = log_{10} (\frac{EW}{λ} )

x = log_{10} [N f (\frac{λ}{5000Å})]

N(Na I) = N(3s) ∗ (1 + \frac{N(3p)}{N(3s)} + \frac{N(4p)}{N(3s)} + ⋯)

= 1.07 × 10^{15} cm^{−2} (1 + 0.05 + 0.0016)

= 1.13 × 10^{15} cm^{−2}

N(Na) = N(Na I) ∗ (1 + \frac{N(Na II)}{N(Na I)})

= 1.13 × 10^{15} cm^{−2} (1 + \frac{2400}{1})

= 2.7 × 10^{18} cm^{−2}

\frac{N(Na)}{N(H)} = \frac{2.7 × 10^{18}}{6.59 × 10^{23}} = 4.1 × 10^{−6}

elemental abundance ≡ 12 + log_{10} (\frac{N(Na)}{N(H)})