/*! 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} Q30P For a helium atom in its ground ... [FREE SOLUTION] | 91影视

91影视

For a helium atom in its ground state, what are quantum numbers (n,l,mI,ms) for the (a) spin-up electron and (b) spin-down electron?

Short Answer

Expert verified
  1. The quantum numbers for the spin-up electron are 1,0,0+12.
  2. The quantum numbers for the spin-down electron are 1,0,0-12.

Step by step solution

01

The given data:

The helium atom is its ground state.

02

Understanding the concept of the quantum number of an atom: 

The set of numbers used to describe the position and power of an electron atom is called quantum numbers. There are four quantum numbers, namely, prime numbers, azimuthal, magnetic numbers, and spin quantum. The stored values of the quantum system are given by quantum numbers.

For every value of the principal quantum number, there are n values of l ranging from 0 to n-1. For every value of the orbital quantum number, there exist 2l+1values of magnetic quantum number ranging from -lto +l. Thus, the maximum value of this magnetic quantum number is equal to the value of the orbital quantum number. Now, every electron has two spin orientations, thus the value of the magnetic quantum number is determined by these spins.

03

(a) Calculation for the quantum numbers of the spin-up electron

When the helium atom is in the ground state, both its electrons are in 1s state. Thus for each electron with quantum numbers n = 1 , l = 0 and ml=0; one of the electrons is in a spin-up quantum state ms=+12and the other electron is in a spin-down ms=-12quantum state.

Hence, the quantum numbers for the spin-up electron are 1,0,0+12.

04

(b) Calculation for the quantum numbers of the spin-down electron

Using the concept of calculations in above part (a), the quantum numbers for the spin-down electron is 1,0,0-12.

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

Show that =1.0610-34J.s=6.5910-16eV.s..

A hypothetical atom has two energy levels, with a transition wavelength between them of . In a particular sample at 300 K,4.01020such atoms are in a state of lower energy. (a) How many atoms are in the upper state, assuming conditions of thermal equilibrium? (b) Suppose, instead, that3.0x1020 of these atoms are 鈥減umped鈥 into the upper state by an external process, with1.01020 atoms remaining in the lower state. What is the maximum energy that could be released by the atoms in a single laser pulse if each atom jumps once between those two states (either via absorption or via stimulated emission)?

What is the wavelength associated with a photon that will induce a transition of an electron spin from parallel to anti-parallel orientation in a magnetic field of magnitude 0.200 T? Assume that l=0.

An atom of uranium has closed 6pand 7ssub shells. Which sub shell has the greater number of electrons?

Comet stimulated emission.When a comet approaches the Sun, the increased warmth evaporates water from the ice on the surface of the comet nucleus, producing a thin atmosphere of water vapor around the nucleus. Sunlight can then dissociate H2Omolecules in the vapor to H atoms and OH molecules. The sunlight can also excite the OH molecules to higher energy levels.

When the comet is still relatively far from the Sun, the sunlight causes equal excitation to the E2and E1levels (Fig. 40-28a). Hence, there is no population inversion between the two levels. However, as the comet approaches the Sun, the excitation to theE1level decreases and population inversion occurs. The reason has to do with one of the many wavelengths鈥攕aid to be Fraunhofer lines鈥攖hat are missing in sunlight because, as the light travels outward through the Sun鈥檚 atmosphere, those particular wavelengths are absorbed by the atmosphere.

As a comet approaches the Sun, the Doppler Effect due to the comet鈥檚 speed relative to the Sun shifts the Fraunhofer lines in wavelength, apparently overlapping one of them with the wavelength required for excitation to the E1level in OH molecules. Population inversion then occurs in those molecules, and they radiate stimulated emission (Fig. 40 28b). For example, as comet Kouhoutek approached the Sun in December 1973 and January 1974, it radiated stimulated emission at about during mid-January. (a) What was the energy difference E2-E1for that emission? (b) In what region of the electromagnetic spectrum was the emission?

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.