/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q6P Using Egs=-79.0 eV  for the g... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Using Egs=-79.0 eV for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, He+, with a single electron orbiting the nucleus; then subtract the two energies.

Short Answer

Expert verified

The first ionization energy is 24.6 eV.

Step by step solution

01

Definition of the hyperfine harmonic oscillator

A harmonic oscillator is a system that experiences a restoring forceFproportionate to the displacement xwhen it is moved from its equilibrium position, where k is a positive constant.

02

The ionization energy

Subtract Helium's ground state energy from its first excited state yields the first ionization energy (energy necessary to remove one electron).

The energy of the Helium ion'sHe+ground state is equal to the energy of the first excited state.

So, calculate the ground state energy ofHe+as:

He+ is a Hydrogen like atom. It contains just one electron but two protons in its nucleus, which indicates it has only one electron.

Hydrogen atom energy,

EH=-13.6 eVZ2n2

ForHydrogen ground state n = 1 , and Z = 1.

For helium, Z = 2, then the expression will be :

EHe=-13.6 eV2212EHe=-54.4 eV

Ground state energy for helium is:

Egs=-79.0 eV

SubtractingEHe andEgs ground state energy as:

Ejoin=-54eV--79eV=24.6eV

Thus, the first ionization energy is 24.6 eV.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose we used a minus sign in our trial wave function (Equation 7.37):ψ=A[ψ0(r1)-ψ0(r2)]): Without doing any new integrals, find F(x) (the analog to Equation ) for this case, and construct the graph. Show that there is no evidence of bonding. (Since the variational principle only gives an upper bound, this doesn't prove that bonding cannot occur for such a state, but it certainly doesn't look promising). Comment: Actually, any function of the form ψ=A[ψ0(r1)+eiϕψ0(r2)]has the desired property that the electron is equally likely to be associated with either proton. However, since the Hamiltonian (Equation 7.35 ) is invariant under the interchange P:r1↔r2 , its eigen functions can be chosen to be simultaneously eigen functions of P . The plus sign (Equation 7.37) goes with the eigenvalue +1 , and the minus sign (Equation 7.52 ) with the eigenvalue -1 ; nothing is to be gained by considering the ostensibly more general case (Equation 7.53), though you're welcome to try it, if you're interested.

Find the best bound on Egsfor the one-dimensional harmonic oscillator using a trial wave function of the form role="math" localid="1656044636654" ψ(x)=Ax2+b2.,where A is determined by normalization and b is an adjustable parameter.

As an explicit example of the method developed inProblem 7.15, consider an electron at rest in a uniform magnetic field B=B2kÁåœfor which the Hamiltonian is (Equation 4.158):

H=-γµþ (4.158).

H0=eBzmSz (7.57).

The eigenspinors, xaarelocalid="1655969802629" xb,andthecorrespondingenergies,EaandEb,aregiven in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction:

{x+,withenergyE+=-γB0ħ/2x-,withenergyE-=+-γB0ħ/2 (4.161).

H'=eBxmSx (7.58).

(a) Find the matrix elements of H′, and confirm that they have the structure of Equation 7.55. What is h?

(b) Using your result inProblem 7.15(b), find the new ground state energy, in second-order perturbation theory.

(c) Using your result inProblem 7.15(c), find the variation principle bound on the ground state energy.

(a) GeneralizeProblem 7.2, using the trial wave function

ψ(x)=A(x2+b2)n,

For arbitrary n. Partial answer: The best value of b is given by

localid="1658300238725" b2=hmÓ¬[n(4n-1)(4n-3)2(2n+1)]1/2

(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form

ψ(x)=Bx(x2+b2)n.

Partial answer: The best value of b is given by

localid="1658300555415" b2=hmÓ¬[n(4n-5)(4n-3)2(2n+1)]1/2.

(c) Notice that the bounds approach the exact energies as n →∞. Why is that? Hint: Plot the trial wave functions for n = 2 , n = 3 , and n = 4, and compare them with the true wave functions (Equations 2.59 and 2.62). To do it analytically, start with the identity

ez=limn→∞(1+zn)nψ0(x)=(mӬπh)1/4e-mӬ2hx2 (2.59).

ψ1(x)=A1a^+ψ0=A12hmӬ(-hddx+mӬx)(mӬπh)1/4e-mӬ2hx2ψ1(x)=A1(mӬπh)1/42mӬhxe-mӬ2hx2(2.62).

Find the best bound on Egsfor the delta-function potentialV(x)=-αδ(x), using a triangular trial function (Equation 7.10, only centered at the origin). This time a is an adjustable parameter.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.