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8.3: Hydrogen Atom

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A hydrogen atom consists of an electron, of charge e and mass me, and a proton, of charge +e and mass mp, moving in the Coulomb potential V(r)=e24πϵ0|r|, where r is the position vector of the electron with respect to the proton. Now, according to the analysis in Section [stwo], this two-body problem can be converted into an equivalent one-body problem. In the latter problem, a particle of mass μ=mempme+mp moves in the central potential V(r)=e24πϵ0r. Note, however, that because me/mp1/1836 the difference between me and μ is very small. Hence, in the following, we shall write neglect this difference entirely.

Writing the wavefunction in the usual form, ψ(r,θ,ϕ)=Rn,l(r)Yl,m(θ,ϕ), it follows from Section 1.2 that the radial function Rn,l(r) satisfies 22me[d2dr2+2rddrl(l+1)r2]Rn,l(e24πϵ0r+E)Rn,l=0. Let r=az, with

a=22me(E)=E0Ea0, where E0 and a0 are defined in Equations ([e9.56]) and ([e9.57]), respectively. Here, it is assumed that E<0, because we are only interested in bound-states of the hydrogen atom. The previous differential equation transforms to [d2dz2+2zddzl(l+1)z2+ζz1]Rn,l=0, where

ζ=2meae24πϵ02=2E0E. Suppose that Rn,l(r)=Z(r/a)exp(r/a)/(r/a). It follows that

[d2dz22ddzl(l+1)z2+ζz]Z=0. We now need to solve the previous differential equation in the domain z=0 to z=, subject to the constraint that Rn,l(r) be square-integrable.

Let us look for a power-law solution of the form

Z(z)=kckzk. Substituting this solution into Equation ([e9.48]), we obtain kck{k(k1)zk22kzk1l(l+1)zk2+ζzk1}=0. Equating the coefficients of zk2 gives the recursion relation

ck[k(k1)l(l+1)]=ck1[2(k1)ζ]. Now, the power series ([e9.49]) must terminate at small k, at some positive value of k, otherwise Z(z) behaves unphysically as z0 [i.e., it yields an Rn,l(r) that is not square integrable as r0]. From the previous recursion relation, this is only possible if [kmin(kmin1)l(l+1)]=0, where the first term in the series is ckminzkmin. There are two possibilities: kmin=l or kmin=l+1. However, the former possibility predicts unphysical behavior of Z(z) at z=0. Thus, we conclude that kmin=l+1. Note that, because Rn,l(r)Z(r/a)/(r/a)(r/a)l at small r, there is a finite probability of finding the electron at the nucleus for an l=0 state, whereas there is zero probability of finding the electron at the nucleus for an l>0 state [i.e., |ψ|2=0 at r=0, except when l=0].

For large values of z, the ratio of successive coefficients in the power series ([e9.49]) is ckck1=2k, according to Equation ([e9.51]). This is the same as the ratio of successive coefficients in the power series k(2z)kk!, which converges to exp(2z). We conclude that Z(z)exp(2z) as z. It thus follows that Rn,l(r)Z(r/a)exp(r/a)/(r/a)exp(r/a)/(r/a) as r. This does not correspond to physically acceptable behavior of the wavefunction, because |ψ|2dV must be finite. The only way in which we can avoid this unphysical behavior is if the power series ([e9.49]) terminates at some maximum value of k. According to the recursion relation ([e9.51]), this is only possible if

ζ2=n, where n is an integer, and the last term in the series is cnzn. Because the first term in the series is cl+1zl+1, it follows that n must be greater than l, otherwise there are no terms in the series at all. Finally, it is clear from Equations ([e9.45]), ([e9.47]), and ([e9.54]) that

E=E0n2 and a=na0, where

E0=mee42(4πϵ0)22=e28πϵ0a0=13.6eV, and

a0=4πϵ02mee2=5.3×1011m. Here, E0 is the energy of so-called ground-state (or lowest energy state) of the hydrogen atom, and the length a0 is known as the Bohr radius. Note that |E0|α2mec2, where α=e2/(4πϵ0c)1/137 is the dimensionless fine-structure constant. The fact that |E0|mec2 is the ultimate justification for our non-relativistic treatment of the hydrogen atom.

We conclude that the wavefunction of a hydrogen atom takes the form

ψn,l,m(r,θ,ϕ)=Rn,l(r)Yl,m(θ,ϕ). Here, the Yl,m(θ,ϕ) are the spherical harmonics (see Section [sharm]), and Rn,l(z=r/a) is the solution of [1z2ddzz2ddzl(l+1)z2+2nz1]Rn,l=0 which varies as zl at small z. Furthermore, the quantum numbers n, l, and m can only take values that satisfy the inequality

|m|l<n, where n is a positive integer, l a non-negative integer, and m an integer.

We expect the stationary states of the hydrogen atom to be orthonormal: that is, ψn,l,mψn,l,mdV=δnnδllδmm, where dV is a volume element, and the integral is over all space. Of course, dV=r2drdΩ, where dΩ is an element of solid angle. Moreover, we already know that the spherical harmonics are orthonormal [see Equation ([spho])]: that is, Yl,mYl,mdΩ=δllδmm. It, thus, follows that the radial wavefunction satisfies the orthonormality constraint 0Rn,lRn,lr2dr=δnn. The first few radial wavefunctions for the hydrogen atom are listed below: R1,0(r)=2a3/20exp(ra0),R2,0(r)=2(2a0)3/2(1r2a0)exp(r2a0),R2,1(r)=13(2a0)3/2ra0exp(r2a0),R3,0(r)=2(3a0)3/2(12r3a0+2r227a20)exp(r3a0),R3,1(r)=429(3a0)3/2ra0(1r6a0)exp(r3a0),R3,2(r)=22275(3a0)3/2(ra0)2exp(r3a0). These functions are illustrated in Figures [coul1] and [coul2].

clipboard_ef0f7b495861294201861c0d4ce84dff9.png

Figure 21: The a0r2|Rn,l(r)|2 plotted as a functions of  r/a0. The solid, short-dashed, and long-dashed curves correspond to n,l=1,0, and 2,0, and 2,1, respectively.

clipboard_e42135b231cd543bdf85a4f90ccda7e7d.png

Figure 22: The a0r2|Rn,l(r)|2 plotted as a functions of  r/a0.The solid, short-dashed, and long-dashed curves correspond to n,l=3,0, and 3,1, and 3,2, respectively. 

Given the (properly normalized) hydrogen wavefunction ([e9.59]), plus our interpretation of |ψ|2 as a probability density, we can calculate rk=0r2+k|Rn,l(r)|2dr, where the angle-brackets denote an expectation value. For instance, it can be demonstrated (after much tedious algebra) that

r2=a20n22[5n2+13l(l+1)],r=a02[3n2l(l+1)],1r=1n2a0,1r2=1(l+1/2)n3a20,1r3=1l(l+1/2)(l+1)n3a30.

According to Equation ([e9.55]), the energy levels of the bound-states of a hydrogen atom only depend on the radial quantum number n. It turns out that this is a special property of a 1/r potential. For a general central potential, V(r), the quantized energy levels of a bound-state depend on both n and l. (See Section 1.3.)

The fact that the energy levels of a hydrogen atom only depend on n, and not on l and m, implies that the energy spectrum of a hydrogen atom is highly degenerate: that is, there are many different states which possess the same energy. According to the inequality ([e9.61]) (and the fact that n, l, and m are integers), for a given value of l, there are 2l+1 different allowed values of m (i.e., l,l+1,,l1,l). Likewise, for a given value of n, there are n different allowed values of l (i.e., 0,1,,n1). Now, all states possessing the same value of n have the same energy (i.e., they are degenerate). Hence, the total number of degenerate states corresponding to a given value of n is 1+3+5++2(n1)+1=n2. Thus, the ground-state (n=1) is not degenerate, the first excited state (n=2) is four-fold degenerate, the second excited state (n=3) is nine-fold degenerate, et cetera (Actually, when we take into account the two spin states of an electron, the degeneracy of the nth energy level becomes 2n2.)

Contributors and Attributions

  • Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)


This page titled 8.3: Hydrogen Atom is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.

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