Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q9P (page 214)

(a) Suppose you put both electrons in a helium atom into the n=2state;

what would the energy of the emitted electron be?

(b) Describe (quantitatively) the spectrum of the helium ion,He+.

Short Answer

Expert verified

(a)The energy of the emitted electron is 2E1-4E1=-2E1=27.2eV

(b)Helium has one electron and it’s a hydrogenise ion with z=2 so the spectrum

is1/λ=4R(1/nf2-1/ni2)

Step by step solution

01

(a) The energy of the emitted electron

The energy of each electron isE=Z2E1/n2=4E1/4=E1=E1=-13.6eV,

so the total initial energy is2×-13.6eV=-27.2eV.

One electron drops to the ground state Z2E1/1=4E1, so the other is left

with 2E1-4E1=-2E1=27.2eV.

02

(b) The spectrum of the helium ion

(b) He+has one electron; it’s a hydrogenise ion with Z = 2, so the spectrum

is 1/λ=4R1/nf2-1/ni2, where R is the hydrogen Rydberg constant, and ni,nfare the

initial and final quantum numbers (1, 2, 3, . . . ).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Thebulk modulus of a substance is the ratio of a small decrease in pressure to the resulting fractional increase in volume:

B=-VdPdV.

Show thatB=(5/3)P, in the free electron gas model, and use your result in Problem 5.16(d) to estimate the bulk modulus of copper. Comment: The observed value is 13.4×1010N/m2, but don’t expect perfect agreement—after all, we’re neglecting all electron–nucleus and electron–electron forces! Actually, it is rather surprising that this calculation comes as close as it does.

Suppose you had three (noninteracting) particles, in thermal equilibrium in a one-dimensional harmonic oscillator potential, with a total energyE=92hω .

(a) If they are distinguishable particles (but all with the same mass), what are the possible occupation-number configurations, and how many distinct (threeparticle) states are there for each one? What is the most probable configuration? If you picked a particle at random and measured its energy, what values might you get, and what is the probability of each one? What is the most probable energy?

(b) Do the same for the case of identical fermions (ignoring spin, as we did in Section 5.4.1).

(c) Do the same for the case of identical bosons (ignoring spin).

(a) Using Equations 5.59 and 5.63, show that the wave function for a particle in the periodic delta-function potential can be written in the form

ψ(X)=C[sinkx+e-ikasina-x]0≤x≤a

(b) There is an exception; At the top of a band where z is an integer multiple ofπyielsψ(x)=0 yields .

Find the correct wave function for the case. Note what happens toψeach delta function.

(a) Calculate<1/r1-r2>for the stateψ0(Equation 5.30). Hint: Dod3r2integral

first, using spherical coordinates, and setting the polar axis alongr1, so

that

ψ0r1,r2=ψ100r1ψ100r2=8πa3e-2r1+r2/a(5.30).

r1-r2=r12+r22-2r1r2cosθ2.

Theθ2integral is easy, but be careful to take the positive root. You’ll have to

break ther2integral into two pieces, one ranging from 0 tor1,the other fromr1to∞

Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it toE0(Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we’re still working with an approximate wave function, so don’t expect perfect agreement.)

E0=8-13.6eV=-109eV(5.31).

In view ofProblem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass.

(a) Find (to two significant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of μ.

E1=-m2h2e24π∞'2=-13.6eV(4.77).

(b) Find the separation in wavelength between the red Balmer lines n=3→n=2for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton).

(c) Find the binding energy of positronium (in which the proton is replaced by a positron—positrons have the same mass as electrons, but opposite charge).

(d) Suppose you wanted to confirm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the “Lyman-α” line n=2→n=1?.

See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Astrophysics

Read Explanation

Engineering Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Fundamentals of Physics

Read Explanation

Electricity and Magnetism

Read Explanation
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.

Sign-up for free