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CHEM 121 Studioby Learn4Less · UBC CHEM 121

CHEM 121 glossary

Rydberg equation

Gives the wavelength of a line of hydrogen (or any one-electron species): 1λ=Z2R∞∣1nf2−1ni2∣\dfrac{1}{\lambda} = Z^2R_\infty\left|\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right| with R∞=1.097×107 m−1R_\infty = 1.097\times10^{7}\ \mathrm{m^{-1}}. It is the energy form ∣ΔE∣=Z2RH∣1nf2−1ni2∣|\Delta E| = Z^2R_H\left|\tfrac{1}{n_f^2} - \tfrac{1}{n_i^2}\right| divided by hchc.

Energy levels of the hydrogen atomn = 1: −2.18 × 10⁻¹⁸ J; n = 2: −5.45 × 10⁻¹⁹ J; n = 3: −2.42 × 10⁻¹⁹ J; n = 4: −1.36 × 10⁻¹⁹ J; n = 5: −8.72 × 10⁻²⁰ J. Arrow from n = 3 to n = 2 (emission).n = ∞0 Jn = 1−2.18 × 10⁻¹⁸ Jn = 2−5.45 × 10⁻¹⁹ Jn = 3−2.42 × 10⁻¹⁹ Jn = 4−1.36 × 10⁻¹⁹ Jn = 5−8.72 × 10⁻²⁰ Jemission

Example

3→23 \to 2 in hydrogen: 1/λ=(1.097×107)(14−19)=1.524×106 m−11/\lambda = (1.097\times10^{7})(\tfrac14 - \tfrac19) = 1.524\times10^{6}\ \mathrm{m^{-1}}, so λ=656\lambda = 656 nm, the red Balmer line.

See alsoBalmer seriesLyman seriesEnergy levelOne-electron species