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Estimate the correction to the ground state energy of hydrogen due to the finite size of the nucleus. Treat the proton as a uniformly charged spherical shell of radius b, so the potential energy of an electron inside the shell is constant:-e2/(4蟺系0b);this isn't very realistic, but it is the simplest model, and it will give us the right order of magnitude. Expand your result in powers of the small parameter, (b / a) whereis the Bohr radius, and keep only the leading term, so your final answer takes the form 螖贰E=A(b/a)n. Your business is to determine the constant Aand the power n. Finally, put in b10-15m(roughly the radius of the proton) and work out the actual number. How does it compare with fine structure and hyperfine structure?

Short Answer

Expert verified

Ground state energy correction is roughly 10-10E1, which is less than fine structure and hyperfine structure correction.

Step by step solution

01

Definition ofhyperfine spliting.

The interaction of the magnetic moments of the electron and proton causes hyperfine splitting, which results in a slightly variable magnetic energy for each spin state.

02

The hyperfine splitting in the ground state of muonic hydrogen.

Inside an evenly charged sphere, the potential is equal to:

V(r)=e24蟺蔚01b-1rH'=-e24蟺蔚01b-1r

Wave function of ground state:

0=e-r/a蟺补2a=4蟺蔚0h2me2

Energy correction of ground state,

E01=<H'0>=-e24蟺蔚01蟺补3e-2r/a1b-1rr2诲谤蝉颈苍蠎诲蠎诲蠁=-e24蟺蔚0a341b0br2e-2r/adr-0bre-2r/adr=-e2蟺蔚0a3a34be-2b/a-2baba+1-1+1-a4a-e-2b/a(a+2b)=-e2蟺蔚0a3a4a2b1+e-2b/a-2b2a2-2ba-1-a+e-2b/a(a+2b)=-e24蟺蔚0a2a2b-e-2b/a2b+2a+a2b-a+e-2b/a(a+2b)=e24蟺蔚0a2a1-ab+e-2b/a1+ab

If ba<<1then e-2b/a1-2ba+124b2a2-168b3a3.

Energy is then equal to:

E01=e24蟺蔚0a1-ab+1+ab1-2ab+2b2a2-4b33a3=e24蟺蔚0a1-ab+1-2ba+2b2a2+ab-2+2ba-4b23a2=e24蟺蔚0a2b2a2-4b23a2=e24蟺蔚0a1a2b23a2

Energy of unperturbed ground state,

E1=-12ae24蟺蔚0E1a=-e24蟺蔚012E01.aE1..a=-43ba2A=-43n=2

If a=510-11then, E01aE1a-510-10which is smaller then correction of fine structure 10-5and hyperfine structure 10-8.

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Most popular questions from this chapter

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?

Analyze the Zeeman effect for the n=3states of hydrogen, in the weak, strong, and intermediate field regimes. Construct a table of energies (analogous to Table 6.2), plot them as functions of the external field (as in Figure 6.12), and check that the intermediate-field results reduce properly in the two limiting cases.

(a) Plugs=0,s=2, and s=3into Kramers' relation (Equation 6.104) to obtain formulas for (r-1),(r),(r-2),and(r3). Note that you could continue indefinitely, to find any positive power.

(b) In the other direction, however, you hit a snag. Put in s=-1, and show that all you get is a relation between role="math" localid="1658216018740" (r-2)and(r-3).

(c) But if you can get (r-2)by some other means, you can apply the Kramers' relation to obtain the rest of the negative powers. Use Equation 6.56(which is derived in Problem 6.33) to determine (r-3) , and check your answer against Equation 6.64.

Prove Kramers' relation:

sn2rs-(2s+1)ars-1+s4[(2l+1)2-s2]a2rs-2=0

Which relates the expectation values of rto three different powers (s,s-1,ands-2),for an electron in the state n/mof hydrogen. Hint: Rewrite the radial equation (Equation) in the form

u''=[l(l+1)r2-2ar+1n2a2]u

And use it to expressrole="math" localid="1658192415441" (ursu'')drin terms of (rs),(rs-1)and(rs-2). Then use integration by parts to reduce the second derivative. Show that (ursu'')dr=-(s/2)(rs-1)and(u'rsu')dr=-[2/s+1](u''rs+1u')dr. Take it from there.

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/acos

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'=-epcos4o0r2

(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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