(a) Using Equation 4.88, work out the first four Laguerre polynomials.

(b) Using Equations 4.86, 4.87, and 4.88, find v(ρ), for the case n=5,I=2.

(c) Find v(ρ)again (for the case role="math" localid="1658315521558" n=5,I=2), but this time get it from the recursion formula (Equation 4.76).

Lq(x)=eqq!(ddx)q(e-x-x9)(4.88)v(ρ)=Ln-2l+1l-1(4.86)Lqp(x)(-1)pddxρLp+q(x)(4.87)cj+1=2(j+l+1-n)(j+1)(j+2l+2)cj(4.76)

Short Answer

Expert verified

(a)TheFirst four Laguerre polynomials are

L0=1L1=1-xL2=2-4x+x2L3=6-18x+9x2-x3

(b)The value ofis v(p) isV(ρ)=21-14ρ+2ρ2

(c)The value of isv(ρ)=c0-23c0ρ+221c0ρ2=c021(21-14ρ+2ρ2)

Step by step solution

01

Definition of the radial wave function

The probability of finding an electron in some finite volume element around a point at a distance of r from the nucleus is given by the radial wave function R(r), which is simply the value of the wave function at some radius r.

02

First four Laguerre polynomials

(a)

The associated Laguerre polynomial is given by equation as

For q=0

L0=exe-xL0=1

Forq=1

L1=exdxd(e-xx)=ex[e-x)-e-xx]=1-x.

for q=2

L2=ex2ddx2exx2=ex2ddx2xe-x-e-x2=ex22e-x-2xe-x+e-xx2-2xe-x=1-2x+12x2

for q=3:

L3=ex6ddx3e-xx3=ex6ddx2-e-xx3+3x2e-x=ex6ddxe-xx3+3x2e-x3x2e-x+6xe-x=ex6-e-xx3+3x2e-x+6x2e-x-12xe-x-6xe-x+6e-x=1-3x+32x2-16x3

Thus,

L0=1L1=1-xL2=2-4x+x2L3=6-18x+9x2-x3

03

Determine v(ρ)

(b)

Need to write vρusing,

v(ρ)=L25(2ρ)L25(x)=(-1)5(dxd)5L7(x)

Need to find L7xas:

L7x=exddx7x7e-x=ex7!ddx67x6e-x-x7e-x=exddx542x5e-x-7x6e-x+x7e-x=exddx4210x4e-x-42x5e-x-84x5e-x+14x6e-x+7x6e-x-x7e-x=exddx22520x2e-x-840+3360x3e-x+840+1260x43-x-252+168x5e-x+228+7x6e-x-x7e-x

=exddx2(2520x2e-x-840+3360x3e-x=840+1260x4e-x-252+168x5e-x+28+7x6e-x-x7e-x)L7x=ex[5040e-x-5040+30240ex-x+15120+37800x2e-xL7x=-12600+8400+8400x3e-x+2100+2100+3150x4e-x-630+252x5e-x+42+7x6e-x-x7e-x]=5040-25280x+52920x2-29400x3+7350x4882x5+49x6-x7

Now, to take the 5th derivative to this to find L25any variable x with power less than 5 will vanish, therefore

localid="1658377935803" L25=-ddx5-882x5+49x6-x7=--8825.4.3.2+49×12x+42x2=60882×2-49×12x+42x2=252042-14x+x2

Thus,

vρ=252042-28ρ+4ρ2=504021-14ρ+2ρ2

04

Calculate v(ρ) using the equation in part (b)

(c)

In this part we will repeat part (b) but using the equations:

v(ρ)=j=0cjpjcj+1=2(j+l+1-n)(j+1)(j+2l+2)cj(4.76)

For n=5andI=2

v52(ρ)=c0+c1ρ+c2ρ2

Where the constants can be determined using the above equation as:

c1=2(3-5)16c0=-23c0.

c2=2(4-5)27c1=-17c1=221c0c3=2(5-5)38

Substitute into, we get

v(ρ)=c0-23c0ρ+221c0ρ2=c021(21-14ρ+2ρ2)

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Most popular questions from this chapter

What is the most probable value of r, in the ground state of hydrogen? (The answer is not zero!) Hint: First you must figure out the probability that the electron would be found between r and r + dr.

Determine the commutator of S2withSZ(1)(whereSS(1)+S(2)) Generalize your result to show that

[S2,S1]=2Ih(S1×S2)

Comment: Because Sz(1)does not commute with S2, we cannot hope to find states that are simultaneous eigenvectors of both. In order to form eigenstates ofS2weneed linear combinations of eigenstates ofSz(1). This is precisely what the Clebsch-Gordan coefficients (in Equation 4.185) do for us, On the other hand, it follows by obvious inference from Equation 4.187that the sumrole="math" localid="1655980965321" S(1)+S(2)does commute withdata-custom-editor="chemistry" S2, which is a special case of something we already knew (see Equation 4.103).

Construct the matrixSrrepresenting the component of spin angular momentum along an arbitrary directionr. Use spherical coordinates, for which

rsinθcosΦı+sinθsinΦø+cosθk [4.154]

Find the eigenvalues and (normalized) eigen spinors ofSr. Answer:

x+(r)=(cosθ/2esinθ/2); x+(r)=(esin(θ/2)-cos(θ/2)) [4.155]

Note: You're always free to multiply by an arbitrary phase factor-say,eiϕ-so your answer may not look exactly the same as mine.

What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus?

  1. First calculate the exact answer, assuming the wave function is correct all the way down tor=0. Let b be the radius of the nucleus.
  2. Expand your result as a power series in the small numbera=2bla, and show that the lowest-order term is the cubic:P(4l3)(bla)3. This should be a suitable approximation, provided thatba(which it is).
  3. Alternatively, we might assume thatψ(r)is essentially constant over the (tiny) volume of the nucleus, so thatP(4l3)πb3lψ(0)l2.Check that you get the same answer this way.
  4. Useb10-15manda05×10-10mto get a numerical estimate forP. Roughly speaking, this represents the fraction of its time that the electron spends inside the nucleus:"

Use equations 4.27 4.28 and 4.32 to construct Y00,Y21Check that they are normalized and orthogonal

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