How much energy in kJ/mole are given off by hydrogen atoms if an electron falls from the n = 10 state to the n = 3 state?

Rydberg formula : 1/λ = R[(1/n₁²) - (1/n₂²)]
and ΔE = Lhc/λ
Then ΔE = LhcR[(1/n₁²) - (1/n₂²)

Avogadro constant, L = 6.022 × 10²³ /mol
Planck constant, h = 6.626 × 10⁻³⁴ J s
Speed of light in vacuum, c = 2.998 × 10⁸ m/s
Rydberg constant, R = 1.097 × 10⁷ m⁻¹

Energy given off, ΔE
= (6.022 × 10²³) × (6.626 × 10⁻³⁴) × (2.998 × 10⁸) × (1.097 × 10⁷) × [(1/3²) - (1/10²)] J/mol
= 132700 J/mol
= 132.7 kJ/mol

132.7