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Calculating the Zeeman Effect``

Problem A

The Ap star BD+0°4535 has a powerful magnetic field of 2.1 T. Calculate the wavelengths of the H[alpha] line caused by Zeeman splitting. (Data from Elkin, Kurtz, Nitschelm, and Unda-Sanzana (2010).)

 

NULL

Hints:

 

Calculate the change in wavelength caused by Zeeman splitting.

 

 

Data

with(ScientificConstants)

 

Electron Mass in kg

me := evalf(Constant(m[e]))

0.9109381882e-30

(2.1)

Electron Charge in J

ec := evalf(Constant(e))

0.1602176462e-18

(2.2)

Speed of Light in m/s

c := evalf(Constant(c))

299792458.

(2.3)

Magnetic Field Strength in TNULL

B := 2.1

 

Wavelength of H`[alpha] Line in m

2.1

(2.4)

lambda[0] := 656.281*10^(-9)

0.6562810000e-6

(2.5)

NULL

 

Useful Equation

Change in Wavelength caused by Zeeman Splitting

 

`Δν` = (1/4)*eB*`πμ`

 

 

Solution

NULL

Calculate the Zeeman split in Hz, using the formula above.

NULL

evalf(ec*B/(4*Pi*me))

0.2939211707e11

(4.1)

 

NULL

`Δν` := 0.2939211707e11

0.2939211707e11

(4.2)

NULL

The Zeeman shift for this magnetic field is 2.939211707*10^10 Hz.

 

The H"[alpha] "line occurs at "lambda[0] = "656.281 nm. This corresponds to

 

nu0 := c/lambda[0]

0.4568050241e15

(4.3)

NULL

Therefore, the three Zeeman frequencies, in Hz, are

 

nu0+`Δν`

0.4568344162e15

(4.4)

nu0

0.4568050241e15

(4.5)

nu0-`Δν`

0.4567756320e15

(4.6)

NULLNULL

The three Zeeman wavelengths, in m, are

 

c/(nu0+`Δν`)

0.6562387757e-6

(4.7)

c/nu0

0.6562810000e-6

(4.8)

c/(nu0-`Δν`)

0.6563232296e-6

(4.9)

NULL

restart

Problem B

Using NASA's Hinode Solar-B telescope, Katsukawa (2011) observed that the iron Fe I line, at 630.25 nm, in the umbra of a large sunspot, was split into two components by Zeeman splitting as shown in the figures below (Katsukawa, 2011).

 

The image shows a sunspot approximately 50,000 kilometres in diameter. The umbra (darkest area) is approximately 15,000 kilometres in diameter. The coloured arrows indicate the magnetic vectors. The letters A, B, C, and D refer to the spectra in the following image.

 

 

The depressions on the right indicate the iron line at 630.25 nanometres. The distance between the two Zeeman wavelenths at the umbra (D) is approximately 28 Ångstroms. Therefore, the line spread is approximately 14 Ångstroms.

 

Find the magnetic field strength of the sunspot's umbra.

 

Hints:

The Fe I line is emitted in a transition from 5D0 to 5P1.

Compute the Landé g factors for both orbitals.

 

Data

Bohr Magneton in ev/G

mu[B] := 5.79*10^(-9)

0.5790000000e-8

(6.1)

Positive Magnetic Spin Number

m[j1] := 1/2

1/2

(6.2)

Negative Magnetic Spin Number

m[j2] := -1/2

-1/2

(6.3)

Iron FE I Line

lambda := 630.25*10^(-9)*Unit('m')

0.6302500000e-6*Units:-Unit('m')

(6.4)

Zeeman Wavelength Shift

`Δλ` := 0.14e-1*10^(-9)*Unit('m')

0.1400000000e-10*Units:-Unit('m')

(6.5)

Planck Constant

h := 6.626*10^(-34)*Unit('J')*Unit('s')

0.6626000000e-33*Units:-Unit('J')*Units:-Unit('s')

(6.6)

Speed of Light

c := 2.997*10^8*Unit('m')/Unit('s')

299700000.0*Units:-Unit('m')/Units:-Unit('s')

(6.7)

 

Useful Equations

Lande g Factor

 

g = 1+(j*(j+1)+s*(s+1)-l*(l+1))/(2*j*(j+1))

Energy from Magnetism at an Orbital

``

`ΔE` = g*m[j]*mu[B]*B

 

Magnetic Field Strength``

B = `Δλ`*h*c/(1.60*lambda^2*`ΔE`*10^(-19)*Unit('J')/Unit('eV'))

``

``

Solution B

``

Following n2S+1Lj for the 5P1 level, j = 1, s = 2, and l = 1:

 

g[2] := 1+(1+1+2*(2+1)-(1+1))/(2*(1+1))

5/2

(8.1)

NULL

For the  5D0 level, j = 0, s = 2, and l = 2. It is clear that the Landé g factor is undefined. We can, therefore, set it to zero in the calculations.

 

 

Find the energy of each level.

 

`ΔE`[pos] := (0*m[j1])*mu[B]*Unit('eV')*B/Unit('G')

0.

(8.2)

``

`ΔE`[neg] := (0*m[j2])*mu[B]*Unit('eV')*B/Unit('G')

0.

(8.3)

``

``

`ΔE`[2*pos] := g[2]*m[j1]*mu[B]*Unit('eV')*B/Unit('G')

0.7237500000e-8*Units:-Unit('eV')*B/Units:-Unit('G')

(8.4)

``

`ΔE`[2*neg] := g[2]*m[j2]*mu[B]*Unit('eV')*B/Unit('G')

-0.7237500000e-8*Units:-Unit('eV')*B/Units:-Unit('G')

(8.5)

 

``

The longest wavelength line (mj = -1/2 → mj = +1/2) is the result of a net energy shift of

 

evalf(`ΔE`[neg]+`ΔE`[2*neg])

-0.7237500000e-8*Units:-Unit('eV')*B/Units:-Unit('G')

(8.6)

evalf(`ΔE`[pos]+`ΔE`[2*pos])

0.7237500000e-8*Units:-Unit('eV')*B/Units:-Unit('G')

(8.7)

 

The total energy difference is

 

`ΔE` := 7.720000000*10^(-9)*Unit('eV')/Unit('G')+7.720000000*10^(-9)*Unit('eV')/Unit('G')

0.1544000000e-7*Units:-Unit('eV')/Units:-Unit('G')

(8.8)

 

``

B = `Δλ`*h*c/(1.60*lambda^2*`ΔE`*10^(-19)*Unit('J')/Unit('eV'))

B = 2833.173838*Units:-Unit('G')

(8.9)

``

The magnetic field is approximately 3000 gauss, which is typical for the umbra of a large sunspot.

``

References

------------------------------------------------------------------------

 

Elkin, V., Kurtz, D., Nitschelm, C. and Unda-Sanzana, E. (2010). The discovery of a 21-kG magnetic field in the Ap star BD+004535. Mon. Not. R. Astron. Soc. 401 (1), L44-L47.

 

Katsukawa, Y. (2011). Measurement of Solar Magnetic Field. Universe of Spectroscopy. http://prc.nao.ac.jp/extra/uos/en/no06/ (accessed: 2016-08-14.)

 

NULL