# H2+ ion (Homework 6-325)

2. H+ 2 ion.

a) Evaluate the direct integral $D = <\psi_0(r)|1/(r-R)|\psi_0(r)> where $|\psi0>i is the ground state wave function of the Hydrogen atom with screening exponent $\zeta= 1$. Hint: use the identity 1/|r−R| = 1/r>, where r> is the larger of r or R so only radial integrals are needed. Compare your result with that of Griffiths Equation (7.47).

b) In atomic units, the ground state energy of $H2+$ for a given internuclear distance $R$ is

$E(R) = 2F(R/a0)$ where $F(x)$ is the function in Griffiths Equation. (7.51). Approximating

$F(x)$ as a parabola $F = F0 +(1/2)F′′(x−x0)^2$, with a minimum at $x = 2.493$ and second

derivative $F′′ = 0.1257$, estimate the vibrational energy $\hbar\omega$ of the protons $\omega= \sqrt{k/μ}$ where $k = d2E/dR2$, and $μ = M_p/2$ is the reduced mass.

a)

$D=<ψ_0 |1/|r ⃗-R ⃗ | | ψ_0> with |ψ_0>=√(1/(πa^3 ))*exp(-r/a)$ ground function in Hydrogen

$D=1/(πa^3 ) ∫_0^∞ 1/(|r ⃗-R ⃗|)*exp(-2r/a)*(r^2 d r) ∮dΩ$ where

$1/|r ⃗-R ⃗ | =1/r_max =$

$=(1/R) \text{ ,for r<R}$

$=(1/r) \text { ,for r≥R}$

$D=4π/(πa^3 ) {∫_0^R r^2/R*exp(-2r/a)d r+∫_R^∞ r*exp(-2r/a)d r}=$

$=4/a^3 {1/R*[a/4(a^2-e^{-2R/a} (a^2+2aR+2R^2)]+a/4 e^{-2R/a} (a+2R)}$

$D=1/R-(1/R+1/a)*exp(-2R/a)$

Equation 7.47 is the same except a scaling factor that comes from a slightly different definition of D in book.

b)

The energy of the ground state is

$E(R/a)=2F(R/a)$ with $F(x)=F_0+(x-x_0 )^2/2*F”(x)$

and $x=R/a$

$E(x)=2F_0+(x-x_0 )^2 *F”(x)$ $(=T+U)$

Since

$k=(d^2 E)/(d x^2 )$

It means

$k=F” (x)=0.1257$

Above $x=R/a$ has no dimension and $F”$ is given in $E_1$ ($13.6 eV$) units. Normally $k$ is given in N/m so that we need to make the following translations:

$F”→F”*13.6 eV$ and $x→R/a=R/(0.53*10^{-10} m)$

Hence

$k=0.1257*13.6*1.6*10^{-19}*1/(0.53*10^{-10} )^2 =97.7 N/m$

Angular frequency is

$ω=√(k/μ)=√(k/(M_p/2))$

So that the vibrational energy is

$E=ℏω=1.055*10^{-34}*√(97.7/(0.5*1.67*10^{-27} ))=3.6*10^{-20} J=0.226 eV$