130_notes.dvi

15.4 Particle in a Sphere* This islike the particle in a boxexcept now the particle is confined to the inside of a sphere of rad ...

We must use the Hankel function of the first type for larger. ρ=kr→iκr κ= √ − 2 μE ̄h^2 Rnℓ=Bh(1)ℓ (iκr) To solve the problem, w ...

equation as we had for 1D odd solution to square well. E -V^0 0 -V 0 +π^2 h^2 /8ma^2 f s if 15.6 Partial Wave Analysis of Scatte ...

Theδℓ(k) is called thephase shiftfor the partial wave of angular momentumℓ. We can compute thedifferential cross sectionfor scat ...

V(r) Scattering from a Spherical Well a r -V 0 E Matching the logarithmic derivative, we get k′ [djℓ(ρ) dρ jℓ(ρ) ] ρ=k′a =k [ Bd ...

With just theℓ= 0 term, the differential scattering cross section is. dσ dΩ → sin^2 (δℓ) k^2 The cross section will have zeros w ...

For the case of a constant potentialV 0 , we definek= √ 2 μ(E−V 0 ) ̄h^2 andρ=kr, and the radial equation becomes. d^2 u(r) dr^2 ...

16 Hydrogen The Hydrogen atom consists of an electron bound to a proton by theCoulomb potential. V(r) =− e^2 r We can generalize ...

and ρ= √ − 8 μE ̄h^2 r. Theprinciple quantum numbernis an integer from 1 to infinity. n= 1, 2 , 3 ,... This principle quantum nu ...

The radial wavefunctions should benormalizedas below. ∫∞ 0 r^2 R∗nℓRnℓdr= 1 See Example 16.4.2:Compute the expected values ofE, ...

...

Auseful integralfor Hydrogen atom calculations is. ∫∞ 0 dx xne−ax= n! an+1 See Example 16.4.2:What is the expectation value of ...

ψ 100 ?* 16.2 The Hydrogen Spectrum The figure shows the transitions between Hydrogen atom states. The ground state of Hydrogen ...

Transitions which changeℓby one unit are strongly preferred, as we will later learn. 16.3 Derivations and Calculations 16.3.1 So ...

Next we look at the equation forlarger. d^2 R dρ^2 − 1 4 R= 0 This can be solved byR=e −ρ (^2) , so we explicitly include this. ...

Thecoefficient of each power ofρmust be zero, so we can derive therecursion relationfor the constantsak. ak+1 ak = k+ℓ+ 1−λ (k+ℓ ...

where ρ= 2 Z na 0 r and the coefficients come from the recursion relation ak+1= k+ℓ+ 1−n (k+ 1)(k+ 2ℓ+ 2) ak. The series termina ...

ak+1= k+ℓ+ 1−n (k+ 1)(k+ 2ℓ+ 2) ak a 1 = 0 + 0 + 1− 2 (0 + 1)(0 + 2(0) + 2) a 0 = − 1 2 a 0 R 20 (r) =C ( 1 − Zr 2 a 0 ) e−Zr/^2 ...

= ̄h 72 i 〈 4 ψ 100 + 3ψ 211 −iψ 210 + √ 10 ψ 21 − 1 |− 3 √ 2 ψ 210 −i √ 2 ψ 211 +i √ 2 ψ 21 − 1 + √ 10 √ 2 ψ 210 〉 = √ 2 ̄h 72 ...

= − ̄h^2 m^2 4 ( Z a 0 ) 3 ∫∞ 0 r^2 e −Zr a 0 ( Z^2 a^20 − 2 Z a 0 r ) e −Zr a (^0) dr = − ̄h^2 m^2 4 ( Z a 0 ) 3 ( Z^2 a^20 2 ( ...