Old approach:


4πj(u,l) = nᵤ (Aᵤₗ + qᵤₗ nₑ) hν.

Let's ignore collisional excitement, making

    4πj(u,l) = n₂ Aᵤₗ hν.

From Allen, Table 4.11,
    
    A[Hβ] = 8.419 × 10⁶ s⁻¹

If H was totally ionized this would be 0, right? Thought I could maybe find a recombination balance to determine a better ratio, but assuming H is not ionized seems to be the approach other students are taking.

From the Boltzman factors computed for hydrogen in hw14, I just reused my spreadsheet to find n₂ = (Z₂/∑Zᵢ * nₜₒₜ) at T=1e4 K:

    n₂ = 0.275 [cm⁻³].

hν = 4.09 × 10⁻¹² [ergs].

4πj(u,l) = n₂ Aᵤₗ hν
            = 0.275 [cm⁻³] 8.419 × 10⁶ [s⁻¹] 4.09 × 10⁻¹² [ergs]
***         = 9.47 × 10⁻⁶ [ergs cm⁻³ s⁻¹].