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
1
Expert's answer
2021-08-23T12:34:24-0400

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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