 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s16/2/b/solution

7000 K

k_{B} T = 0.603 eV

and

C_{12} = 1.415 \times 1 0^{- 13} m^{3} s^{- 1}, n_{e} = \frac{0.007 A _{21}}{C _{12}} = 1.13 \times 1 0^{5} m^{- 3} = 0.113 c m^{- 3} .

The

2 p

ground term has fine-structure levels

J = 1/2

and

J = 3/2

g_{1} = 2

and

g_{2} = 4

C_{21} = C_{12} (g_{1} / g_{2}) e^{Δ E / (k_{B} T)} = 7.17 \times 1 0^{- 14} m^{3} s^{- 1}

. The critical density is therefore

n_{crit} = A_{21} / C_{21} = 3.19 \times 1 0^{7} m^{- 3} ≃ 32 c m^{- 3} .

With collisions and spontaneous decay both retained,

N_{2} / N_{1} = n_{e} C_{12} / (A_{21} + n_{e} C_{21})

N_{2} / N_{1} \to 2 e^{- Δ E / (k_{B} T)} ≃ 1.97

, requires

n_{e} ≫ n_{crit}

, not merely equality. For example

n_{e} ≳ 3 \times 1 0^{8} m^{- 3}

brings the ratio within about ten per cent of that limit.

 New to topics? Read the docs here!