Question #228713

The calcium line of wavelength λ = 422.673 nm (P → S) exhibits normal Zeeman
splitting when placed in uniform magnetic field of 4 webers/meter2
. Calculate the
wavelength of three components of normal Zeeman pattern and the separation between
them

Expert's answer

Splitting between energy level in the presence of magnetic field is given by

hcλ2Δλ=±μBBμB=9.27×1027  Am2B=4  Wb/m2Δλ=±μBBλ2hcλ=422.673  nmΔλ=±9.27×4×(422.673×109)26.626×3×1026=±3.332×1011=±0.033  nm\frac{hc}{λ^2}Δλ=±\mu_BB \\ \mu_B = 9.27 \times 10^{-27} \;A \cdot m^2 \\ B=4 \;Wb/m^2 \\ Δλ = ± \frac{\mu_B B λ^2}{hc} \\ λ = 422.673 \; nm \\ Δλ = ± \frac{9.27 \times 4 \times (422.673 \times 10^{-9})^2}{6.626 \times 3 \times 10^{-26}} \\ = ± 3.332 \times 10^{-11} \\ = ± 0.033 \;nm

The wavelength of three components of normal Zeeman pattern

λ, λ+Δλ, λ-Δλ = 422.673 nm, 722.706 nm, 422.640 nm

Splitting

Δλ=0.033 nm


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