/*! 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} Q83CE Exercise 81 obtained formulas fo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Q83CE (page 284)

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass m1, whilem2is the mass of the orbiting negative charge. In positronium, an electron orbits a single positive charge, as in hydrogen, but one whose mass is the same as that of the electron -- a positron. Obtain numerical values of the ground state energy and 鈥淏ohr radius鈥� of positronium.

Short Answer

Expert verified

(a) The ground State energy of positronium is -6.8 eV .

(b) Bohr radius of positronium is 0.106 nm .

Step by step solution

01

Energy of ground-state of positronium

As you know from that, the ground state energy is,

Eground=Z2渭mE1n2

Where, zis the atomic number, 渭is the Reduced mass, n is the principal quantum number, m is the mass, E1is the Energy of ground state of hydrogen atom.

You also know that the electron and positron have same masses.

Hence, the reduced mass will be half of the mass of electron

Eground=Z2渭mE1n2=1212mmE112=-6.8eV

Hence, ground state energy of the positronium is -6.8 eV .

02

Bohr Radius of the positronium

As you know that,

The Bohr鈥檚 Radius

rn=mz渭a0

Where, a0is radius of hydrogen atom

If the reduced mass will be half of the mass of electron,

rn=mZ渭a0=mZ12ma0=0.106nm

Hence, Bohr Radius of the positronium is 0.106 nm.

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 91影视!

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

Exercise 80 discusses the idea of reduced mass. When two objects move under the influence of their mutual force alone, we can treat the relative motion as a one particle system of mass 渭=m1v2/(m1+m2). Among other things, this allows us to account for the fact that the nucleus in a hydrogen like atom isn鈥檛 perfectly stationary, but in fact also orbits the centre of mass. Suppose that due to Coulomb attraction, an object of mass m2and charge -eorbits an object of mass m1 and charge +Ze . By appropriate substitution into formulas given in the chapter, show that (a) the allowed energies are Z2渭mE1n2, where is the hydrogen ground state, and (b) the 鈥淏ohr Radius鈥� for this system is m锄渭a0 ,where a0is the hydrogen Bohr radius.

A simplified approach to the question of how lis related to angular momentum 鈥� due to P. W. Milonni and Richard Feynman 鈥� can be stated as follows: If can take on only those values mlh, whereml=0,卤1,鈥�鈥�卤l , then its square is allowed only valuesml2h2, and the average of localid="1659178449093" l2should be the sum of its allowed values divided by the number of values,2l+1 , because there really is no preferred direction in space, the averages of Lx2andLy2should be the same, and sum of all three should give the average of role="math" localid="1659178641655" L2. Given the sumrole="math" localid="1659178770040" 鈭�1Sn2=N(N+1)(2N+1)/6, show that these arguments, the average of L2 should be l(l+1)h2.

Roughly, how does the size of a triply ionized beryllium ion compare with hydrogen?

We have noted that for a given energy, as lincreases, the motion is more like a circle at a constant radius, with the rotational energy increasing as the radial energy correspondingly decreases. But is the radial kinetic energy 0 for the largest lvalues? Calculate the ratio of expectation values, radial energy to rotational energy, for the(n,l,mt)=(2.1,+1)state. Use the operators

KE铃�rad=-h22m1r2鈭�鈭�r(r鈭�鈭�r)KE铃�rad=h2l(l+1)2mr2

Which we deduce from equation (7-30).

Question: Section 7.5 argues that knowing all three components of would violate the uncertainty principle. Knowing its magnitude and one component does not. What about knowing its magnitude and two components? Would be left any freedom at all and if so, do you think it would be enough to satisfy the uncertainly principle?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Energy Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Kinematics Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Fields in Physics

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.