(a) Construct the wave function for hydrogen in the state n=4,I=3,m=3. Express your answer as a function of the spherical coordinates r,θandϕ.

(b) Find the expectation value of role="math" localid="1658391074946" rin this state. (As always, look up any nontrivial integrals.)

(c) If you could somehow measure the observable Lx2+Ly2on an atom in this state, what value (or values) could you get, and what is the probability of each?

Short Answer

Expert verified

(a) The wave function for hydrogen in the given states is

ψ433=-16144πa9r3e-r/4asin3(θ)e3iϕ.

(b) The expectation value in the state is 18a.

(c ) The value is3ħ2 .

Step by step solution

01

Define the wave function

The location of an electron at a specific place in space (defined by its x, y, and z coordinates) and the amplitude of its wave, which corresponds to its energy, are related by a mathematical function known as a wave function,ψ.

02

Step 2: (a) Construct the wave function for hydrogen

The equation for the spatial wave function of a hydrogen atom ,

ψn/m=2na3n-I-1!2nn+I!3e-r/na2rnaILn-I-12I+12rnaYImθ,ϕ

HereL is an associated Laguerre polynomial andY is a spherical harmonic, and they are given as follow:

Lpqx=c0j=0p-1jp+q!p-j!q+j!j!xjYImθ,ϕ=2I+1I-m!4πI+m!-1/2eimoPImcosθ

HerePImis the associated Legendre function:

Plm(x)=(1-x2)m/2ddx|m|Pl(x)

And,

Pl(x)=12ll!ddxI(x2-1)l

For n=4,I= 3 and m =3, determine Y33. To find it construct YII

Yll=(-1)I(2l+1)4π12I!eilϕPll(cos(θ))

From (1), writePIIas:

Pll(x)=(1-x2)1/2ddxIPl(x)

Replace from (5) with PI

Pll(x)=12ll!(1-x2)I/2ddx2I(x2-1)l

but (x2-1)l=x2I+, and the remaining term has a power less than 2I.So, when differentiate (x2-1)l,2ltimes all the terms vanishes except the first term with the power of , thus:

Pll(x)=12ll!(1-x2)I/2ddx2x2I

Now,

ddxnxn=n!

Hence:

Pll=(2l)!2ll!(1-x2)I/2

Next for x=cos(θ),1-x2=sin2(θ):

Pll=(2l)!2ll!sinl(θ)

So:

YII=-1I2I+14π2I!eiIϕ2I!2II!siniθ=-1I2I!2I+14πeiIϕ12II!sinIθ=1I!2I+1!4π-12eiϕsinθI

Again, forI=3,

y33=-3564π12sin3θe3iϕ

Also, for n=4,l=3and m=3, use L07(x)=7!=5040.Substitute L07(x)andY33into the overall formula (1),

localid="1658397501310" ψ433=12a316×50403e-r/4ar2a35040-3564π12sin3θe3iϕ=-16144πa9r3e-r/4asin3θe3iϕψ433

Therefore, the wave function for hydrogen in the given states as a function of the spherical coordinatesr,θand=-16144πa9r3e-r/4asin3θe3iϕψ433is .

03

(b) Determine the expectation value of r

Evaluate the expectation value ofr, that is r, as:

localid="1658403974750" r=rψ2d3r=161442πa9rr6e-r/2asin6θr2sinθdrdθdϕ=161442πa90r9e-r/2a0πθdθ02πdϕ=161442πa99!2a1021.4.63.5.72π

Further evaluate and get,

=18ar=18a

Thus, the expectation value of r in the state is 18a.

04

(c) Find the probability.

Assume thatbe an eigen function of the operator with eigen value of l(l+1)2,forI=3

Thus:

L2=II+1ħ2=12ħ2

Supposerole="math" localid="1658398355830" ψ433be an eigen function of the operator Lzwith eigen value of mtform=3,:

Lz=3

Hence:

role="math" localid="1658398913271" Lx2+Ly2=L2-Lz2=12ħ2-9ħ2=3ħ2=3ħ2

Thus, the required value is3ħ2

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Most popular questions from this chapter

Coincident spectral lines. 43According to the Rydberg formula (Equation 4.93) the wavelength of a line in the hydrogen spectrum is determined by the principal quantum numbers of the initial and final states. Find two distinct pairs{ni,nf} that yield the same λ. For example,role="math" localid="1656311200820" {6851,6409} and{15283,11687}will do it, but you're not allowed to use those!

(a)Derive Equation 4.131 from Equation 4.130. Hint: Use a test function; otherwise you're likely to drop some terms.

(b)Derive Equation 4.132 from Equations 4.129 and 4.131 .Hint : Use Equation 4.112.

(a) Apply S-to|10 (Equation4.177 ), and confirm that you get 2|1-1

(b) Apply S±to[00 (Equation 4.178), and confirm that you get zero.

(c) Show that |11 and |1-1 (Equation 4.177) are eigenstates of S2, with the appropriate eigenvalue

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:"

[Refer to. Problem 4.59for background.] Suppose A=B02(X^-yI^) andφ=Kz2, where B0 and Kare constants.

(a) Find the fields E and B.

(b) Find the allowed energies, for a particle of mass m and charge q , in these fields, Answer: E(n1,n2)=(n1+12)ħω1+(n2+12)ħω2,(n1,n2=0,1,2,...)whereω1qB0/mandω22qK/m. Comment: If K=0this is the quantum analog to cyclotron motion;ω1 is the classical cyclotron frequency, and it's a free particle in the z direction. The allowed energies,(n1+12)ħω1, are called Landau Levels.

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