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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=14蟺蔚0.emc2r3L

Here,0is the permittivity of the free space with value role="math" localid="1658138846877" 8.910-10C2/N.m2,e is the charge on electron with value 1.610-19C,m, is the mass of the electron with value 9.110-31kg,c is the speed of light with value 3108m/s,ris the Bohr鈥檚 radius with value 0.5310-10m, and L=which is Planck鈥檚 constant with value 1.0510-34J.s.

02

Determination of the internal magnetic field of the hydrogen atom

Assume r=a that is Bohr鈥檚 radius, and L=.

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

B=14蟺蔚0.emec2a3=148.910-10C2/N.m21N.m1J.1.610-19C1.0510-34J.s9.110-31kg3108m/s20.5310-10m=12C.m/s1T1C.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

Two identical spin-zero bosons are placed in an infinite square well (Equation 2.19). They interact weakly with one another, via the potential

V(x1,x2)=-aV0(x1-x2). (2.19).

(where V0is a constant with the dimensions of energy, and a is the width of the well).

(a)First, ignoring the interaction between the particles, find the ground state and the first excited state鈥攂oth the wave functions and the associated energies.

(b) Use first-order perturbation theory to estimate the effect of the particle鈥 particle interaction on the energies of the ground state and the first excited state.

Suppose the Hamiltonian H, for a particular quantum system, is a function of some parameter let En()and n()be the eigen values and

Eigen functions of. The Feynman-Hellmann theorem22states that

En=(nHn)

(Assuming either that Enis nondegenerate, or-if degenerate-that the n's are the "good" linear combinations of the degenerate Eigen functions).

(a) Prove the Feynman-Hellmann theorem. Hint: Use Equation 6.9.

(b) Apply it to the one-dimensional harmonic oscillator,(i)using =(this yields a formula for the expectation value of V), (II)using =(this yields (T)),and (iii)using =m(this yields a relation between (T)and (V)). Compare your answers to Problem 2.12, and the virial theorem predictions (Problem 3.31).

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]

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

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]

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

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

Without the Coulomb interaction it would have been E0=0, where 0=k/m. Assuming that, show that

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).

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

Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as inProblem 2.46).

(a) Show that the stationary states can be written in the formn(x)=1Le2inx/L,(-L/2<x<L/2),

wheren=0,1,2,....and the allowed energies areEn=2mnL2.Notice that with the exception of the ground state (n = 0 ) 鈥 are all doubly degenerate.

(b) Now suppose we introduce the perturbation,H'=-V0e-x2/a2where aLa. (This puts a little 鈥渄imple鈥 in the potential at x = 0, as though we bent the wire slightly to make a 鈥渢rap鈥.) Find the first-order correction to En, using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that aLato extend the limits from L/2toafter all, H鈥 is essentially zero outside -a<x<a.

E1=12Waa+WbbWaa-Wbb2+4Wab2(6.27).

(c) What are the 鈥済ood鈥 linear combinations ofnand-n, for this problem? Show that with these states you get the first-order correction using Equation 6.9.

En'=n0H'n0(6.9).

(d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous Eigenstates ofH0and A are precisely the ones you used in (c).

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