Use Equation 6.59 to estimate the internal field in hydrogen, and characterize quantitatively a "strong" and "weak" Zeeman field.

Short Answer

Expert verified

The value of internal magnetic field is 12 T .

The strong and weak Zeeman field is characterized as:

BextBintstrong Zeeman field

BextBintweak Zeeman field

Step by step solution

01

Expression for the internal magnetic field

The expression for the internal magnetic field in the hydrogen atom is given as follows,

B=14πε0.emc2r3L

Here,ε0is the permittivity of the free space with value role="math" localid="1658138846877" 8.9×10-10C2/N.m2,e is the charge on electron with value 1.6×10-19C,m, is the mass of the electron with value 9.1×10-31kg,c is the speed of light with value 3×108m/s,ris the Bohr’s radius with value 0.53×10-10m, and L=ħwhich is Planck’s constant with value 1.05×10-34J.s.

02

Determination of the internal magnetic field of the hydrogen atom

Assume r=a that is Bohr’s radius, and L=ħ.

Substitute the values in the expression for the internal magnetic field in the hydrogen atom.

B=14πε0.emec2a3ħ=14π8.9×10-10C2/N.m2×1N.m1J.1.6×10-19C1.05×10-34J.s9.1×10-31kg3×108m/s20.53×10-10m=12C.m/s×1T1C.m/s=12T

03

Quantitative characterization of strong and weak Zeeman field

It is known that the strong Zeeman field is Bext>>10Tand the weak Zeeman field is Bext10T. So,BextBintandBextBint .

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Most popular questions from this chapter

(a) Find the second-order correction to the energies(En2)for the potential in Problem 6.1. Comment: You can sum the series explicitly, obtaining -for odd n.

(b) Calculate the second-order correction to the ground state energy(E02)for the potential in Problem 6.2. Check that your result is consistent with the exact solution.

When an atom is placed in a uniform external electric field ,the energy levels are shifted-a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyse the Stark effect for the n=1 and n=2 states of hydrogen. Let the field point in the z direction, so the potential energy of the electron is

H's=eEextz=eEextrcosθ

Treat this as a perturbation on the Bohr Hamiltonian (Equation 6.42). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.)

(a) Show that the ground state energy is not affected by this perturbation, in first order.

(b) The first excited state is 4-fold degenerate: Y200,Y211,Y210,Y200,Y21-1Using degenerate perturbation theory, determine the first order corrections to the energy. Into how many levels does E2 split?

(c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment (pe=-er) in each of these "good" states.Notice that the results are independent of the applied field-evidently hydrogen in its first excited state can carry a permanent electric dipole moment.

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'=14πϵ0(e2R-e2R-x1-e2R+x2+e2R-x1+x2 [6.97]

(a) Explain Equation6.97. Assuming that localid="1658203563220" |x1| and |x2|are both much less than, show that

localid="1658203513972" H'-e2x1x22πϵ0R3 [6.98]

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

H=[12mp+2+12(k-e22πϵ0R3x+2]+[+12mp-2+12(k+e22πϵ0R3x-2] [6.99]

under the change of variables

x±12(x1±x2) Which entails p±=12(p1±p2) [6.100]

(c) The ground state energy for this Hamiltonian is evidently

E=12(ω++ω-) Where ω±=k(e2/2πϵ0R3)m [6.101]

Without the Coulomb interaction it would have been E0=ħω0, where ω0=k/m. Assuming that, show that

ΔVE-E0-8m2ω03(e22πϵ0)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 ψn1(x1)ψn2(x2), where ψn(x)is a one-particle oscillator wave function with mass mand spring constant k;ΔVis the second-order correction to the ground state energy, for the perturbation in Equation 6.98 (notice that the first-order correction is zero).

For the harmonic oscillator[Vx=1/2kx2], the allowed energies areEN=(n+1/2)ħω,(n=0.1.2,..),whererole="math" localid="1656044150836" ω=k/mis the classical frequency. Now suppose the spring constant increases slightly:k(1+ο')k(Perhaps we cool the spring, so it becomes less flexible.)

(a) Find the exact new energies (trivial, in this case). Expand your formula as a power series inο,, up to second order.

(b) Now calculate the first-order perturbation in the energy, using Equation 6.9. What ishere? Compare your result with part (a).

Hint: It is not necessary - in fact, it is not permitted - to calculate a single integral in doing this problem.

Question: Sometimes it is possible to solve Equation 6.10 directly, without having to expand in terms of the unperturbed wave functions (Equation 6.11). Here are two particularly nice examples.

(a) Stark effect in the ground state of hydrogen.

(i) Find the first-order correction to the ground state of hydrogen in the presence of a uniform external electric fieldEext (the Stark effect-see Problem 6.36). Hint: Try a solution of the form

(A+Br+Cr2)e-r/acosθ

your problem is to find the constants , and C that solve Equation 6.10.

(ii) Use Equation to determine the second-order correction to the ground state energy (the first-order correction is zero, as you found in Problem 6.36(a)). Answer:-m(3a2eEext/2)2 .

(b) If the proton had an electric dipole moment p the potential energy of the electron in hydrogen would be perturbed in the amount

H'=-epcosθ4π̀o0r2

(i) Solve Equation 6.10 for the first-order correction to the ground state wave function.

(ii) Show that the total electric dipole moment of the atom is (surprisingly) zero, to this order.

(iii) Use Equation 6.14 to determine the second-order correction to the ground state energy. What is the first-order correction?

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