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Consider the (eight) n=2states,|2lmlms.Find the energy of each state, under strong-field Zeeman splitting. Express each answer as the sum of three terms: the Bohr energy, the fine-structure (proportional toa2), and the Zeeman contribution (proportional toBBext.). If you ignore fine structure altogether, how many distinct levels are there, and what are their degeneracies?

Short Answer

Expert verified

Only five of the eight distinct energies remain when fine structure is ignored: There are three degenerate energies.

Step by step solution

01

Definition of Zeeman splitting.

When radiation (such as light) originates in a magnetic field, Zeeman splitting occurs, which is the splitting of a single spectral line into two or more lines of different frequencies.

02

Step2: Derivation.

For n=2, there are eight different states |2lmlms:

i)l=0ml=0ms=-12,ii)l=0ml=0ms=12,iii)l=1ml=-1ms=-12,iv)l=1ml=-1ms=12,v)l=1ml=0ms=-12,vi)l=1ml=0ms=12,vii)l=1ml=1ms=-12,viii)l=1ml=1ms=1/2i)l=0ml=0ms=-12,ii)l=0ml=0ms=12,iii)l=1ml=-1ms=-12,iv)l=1ml=-1ms=12,v)l=1ml=0ms=-12,vi)l=1ml=0ms=12,vii)l=1ml=1ms=-12,viii)l=1ml=1ms=12

Total energy is equal to:

E=-13.6eVn2+BBext(ml+2ms)+13.6eVn3234n-l(l+1)-mlmsll+12(l+1)

role="math" localid="1656064856692" i)E=-13.6eVn2-BBext+13.6eVn3234n-1.ii)E=-13.6eVn2+BBext+13.6eVn3234n-1.iii)E=-13.6eVn2-2BBext+13.6eVn3234n-32.iv)E=-13.6eVn2+13.6eVn3234n-56.v)E=-13.6eVn2-BBext+13.6eVn3234n-23.vi)E=-13.6eVn2+BBext+13.6eVn3234n-23.vii)E=-13.6eVn2+2BBext+13.6eVn3234n-56.viii)E=-13.6eVn2+13.6eVn3234n-12.

If fine structure is ignored,a=0.

Therefore, the following energies:

E=-13.6eVn2-BBextDegree of degeneracy: 2

E=-13.6eVn2+BBextDegree of degeneracy: 2

E=-13.6eVn2Degree of degeneracy: 2

E=-13.6eVn2-2BBextDegree of degeneracy: 1

E=-13.6eVn2+2BBextDegree of degeneracy: 1

Now only there are five distinct energies instead of eight.

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Most popular questions from this chapter

Starting with Equation 6.80, and using Equations 6.57, 6.61, 6.64, and 6.81, derive Equation 6.82.

Consider the (eight) n=2states, |2ljmj. Find the energy of each state, under weak-field Zeeman splitting, and construct a diagram like Figure 6.11 to show how the energies evolve asBext increases. Label each line clearly, and indicate its slope.

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Work out the matrix elements of HZ'andHfs'construct the W matrix given in the text, for n = 2.

Van der Waals interaction. Consider two atoms a distanceapart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact a weak attraction. To model this system, picture each atom as an electron (mass m , charge -e ) attached by a spring (spring constant k ) to the nucleus (charge +e ), as in Figure. We'll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is

H0=12mp12+12kx12+12mp22+12kx22[6.96]

The Coulomb interaction between the atoms is

H'=14蟺系0(e2R-e2R-x1-e2R+x2+e2R-x1+x2 [6.97]

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localid="1658203513972" H'-e2x1x220R3 [6.98]

(b) Show that the total Hamiltonian (Equationplus Equation) separates into two harmonic oscillator Hamiltonians:

H=[12mp+2+12(k-e220R3x+2]+[+12mp-2+12(k+e220R3x-2] [6.99]

under the change of variables

x12(x1x2) Which entails p=12(p1p2) [6.100]

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E=12(++-) Where =k(e2/20R3)m [6.101]

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VE-E0-8m203(e220)21R6. [6.102]

Conclusion: There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. This is the van der Waals interaction between two neutral atoms.

(d) Now do the same calculation using second-order perturbation theory. Hint: The unperturbed states are of the form n1(x1)n2(x2), where n(x)is a one-particle oscillator wave function with mass mand spring constant k;Vis the second-order correction to the ground state energy, for the perturbation in Equation 6.98 (notice that the first-order correction is zero).

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