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Asked in RS Academy GUJCET booklet · Ranges, production and detection
Given: λ≈5890 A, and 1 A=10⁻¹⁰ m.
Idea: convert to nanometres and compare with the visible band, which runs from about 400 to 700 nm.
5890 A=589×10⁻⁹ m=589 nm.
That sits squarely inside the visible range, in the yellow -- it is the familiar yellow of a sodium street lamp.
So the doublet belongs to the visible light region.
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