Question

Ratio of intensities of D1 and D2 lines of sodium

Answer 1

Answer:

Assuming that the plasma is in local thermal equilibrium it is expected that the intensity ratio of sodium doublet components equals the ratio of the statistical weights of the atomic sublevels corresponding to the electronic transitions (D1: 32P1/2–32S1/2, D2: 32P3/2–32S1/2) i.e. 2.

Answer 2

Answer:

The ratio intensities of D₁ and D₂ lines of Sodium at high temperature is 1:2

Explanation:

• The ratio of D₁ and  D₂ are not constant it may vary from 2 to 1.2 . The density of sodium atom was known the ratio of intensities of the emission lines.
• The intensity of the lines can be reduced by the ratio if the size of the emitting region is less than the width of the slit and is independent of "The Slit filling Factor"

The Electronic transition for the line  D₁ and  D₂ is given by:

$D_{1} = P^{2} _{\frac{1}{2} } - > S^{2} _{\frac{1}{2} } \\\\D_{2} = P^{2} _{\frac{1}{2} } - > S^{2} _{\frac{1}{2} }$

The ratio of Intensity of the line  D₁ and  D₂ is given by:

$\frac{I (D_{2}) }{I(D_{1}) } = \frac{2 J_{2} + 1 }{2J_{1} +1 }$    .....................................(1)

$J_{1} = \frac{1}{2}\\ \\ and\\\\J_{2} = \frac{3}{2}$

Substitute these values in equation (1) we get

$\frac{I (D_{2}) }{I(D_{1}) } = \frac{2 (\frac{3}{2}) + 1 }{2(\frac{1}{2}) +1 }$

$\frac{I (D_{2}) }{I(D_{1}) } = \frac{4}{2} \\ \\\frac{I (D_{2}) }{I(D_{1}) } = \frac{1}{2}$

Therefore, D₁ : D₂ = 1 : 2

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