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Chapter 7: Q47E (page 282)

Question: Show that the angular normalization constant in Table 7.3 for the case (l,ml)=(1,0) is correct.

Short Answer

Expert verified

Answer

It has been proved that the normalization for the case (l,ml)=(1,0)is correct.

Step by step solution

01

Given data

The wave function corresponding to l,ml=1,0 is,

Θl,mlθΦmlϕ=34πcosθ

02

Normalization

TheangularpartoftheHydrogenatomwavefunctionshouldsatisfytheflowingcondition,∫0πΘl,mlθ22πsinθdθ=1 .....(I)

03

Determining whether the given normalization constant is correct

In the given wave function,

Θl,mlθ=34πcosθ

Check equation (I) as,

=34π×2π∫0πcos2θsinθdθ=32∫0πcos2θsinθdθ

Let us assume,

cosθ=z-sinθdθ=dz

Then the integral becomes,

=32∫1-1z2-dz=32∫-11z2dz=32z33-11=32×23=1

Thus the normalization is correct.

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Most popular questions from this chapter

A simplified approach to the question of how lis related to angular momentum – due to P. W. Milonni and Richard Feynman – can be stated as follows: If can take on only those values mlh, whereml=0,±1,……±l , then its square is allowed only valuesml2h2, and the average of localid="1659178449093" l2should be the sum of its allowed values divided by the number of values,2l+1 , because there really is no preferred direction in space, the averages of Lx2andLy2should be the same, and sum of all three should give the average of role="math" localid="1659178641655" L2. Given the sumrole="math" localid="1659178770040" ∑1Sn2=N(N+1)(2N+1)/6, show that these arguments, the average of L2 should be l(l+1)h2.

Residents of flatworld-a two-dimensional world far, far away-have it easy. Although quantum mechanics of course applies in their world, the equations they must solve to understand atomic energy levels involve only two dimensions. In particular, the Schrodinger equation for the one-electron flatrogen atom is

-h2m1r∂∂r(r∂∂r)ψ(r,θ)-h2m1r2∂2∂θ2ψ(r,θ)+U(r)ψ(r,θ)=Eψ(r,θ)

(a) Separate variables by trying a solution of the form ψ(r,θ)=R(r)⊝(θ), then dividing byR(r)⊝(θ) . Show that the θequation can be written

d2dθ2⊝(θ)=C⊝(θ)

Here,(C) is the separation constant.

(b) To be physically acceptable,⊝(θ) must be continuous, which, since it involves rotation about an axis, means that it must be periodic. What must be the sign of C ?

(c) Show that a complex exponential is an acceptable solution for⊝(θ) .

(d) Imposing the periodicity condition find allowed values ofC .

(e) What property is quantized according of C .

(f) Obtain the radial equation.

(g) Given thatU(r)=-b/r , show that a function of the formR(r)=er/a is a solution but only if C certain one of it, allowed values.

(h) Determine the value of a , and thus find the ground-state energy and wave function of flatrogen atom.

For the cubic 3D infinite well wave function

ψ(x,y,z)=AsinnxπxLsinnyπyLsinnzπzLshow that the correct normalization constant isA=(2/L)3/2.

An electron in a hydrogen atom is in the (n,l,ml) = (2,1,0) state.

(a) Calculate the probability that it would be found within 60 degrees of z-axis, irrespective of radius.

(b) Calculate the probability that it would be found between r = 2a0 and r = 6a0, irrespective of angle.

(c) What is the probability that it would be found within 60 degrees of the z-axis and between r = 2a0 and r = 6a0?

In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,λ=2L/n , where is the length of its home. Given that λ=h/p, we would have p=nh/2L, and the kinetic energy, p2/2m, would thus be n2h2/8mL2. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: E=U. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?
See all solutions

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