--%>

Electron Spin

The total angular momentum of an atom includes an electron spin component as well as an orbital component.

The orbital motion of each electron of an atom contributes to the angular momentum of the atom, as described earlier. An additional angular momentum contribution comes from the “spin of the electrons.”

The direct experimental demonstration of an electron feature that is described as spin angular momentum was given by the atomic beam studies of O. Stern and W. Gerlach. In the original experiments, a beam of silver atoms was passed through a magnetic field. The result was a splitting of the atom into two components. Thus, when a directional field is composed, two different states of silver atoms can be recognized.

The lowest energy electronic state of silver atoms consists of inner shells of electrons and a single outer shell electron in atom in an s orbital. No additional states should be developed when a directional field is applied to this spherically symmetric, zero angular momentum atom. The Stern-Gerlach results supported the idea that the silver atoms have an angular momentum of ½ h/ (2∏), or 1/2 h, which results from the intrinsic angular momentum of the electron. The magnetic field distinguishes those atoms with a spin angular momentum directed with and opposed to the field. If the electron spins quantum number s has a value of 1/2, jection of the spin angular momentum along an imposed direction is given by m2, h, where m2 = +1/2 or – ½.

In describing the electronic makeup of atoms, we use angular momentum to characterize the atomic states. From the above equations the orbital angular momentum contribution of an electron is √l (l + 1) h, where l = 0, 1, 2 …  now there is, in addition, an electronic spin angular momentum contributions are used  to describe the states of many electron atoms.

   Related Questions in Chemistry

  • Q : Short note on the function of

    Write down a short note on the function of mitochondria?

  • Q : Concentration factor affected by

    Can someone please help me in getting through this problem. Which of the given concentration factor is affected by the change in temperature: (1) Molarity (2) Molality (3) Mole fraction (4) Weight fraction

  • Q : Explain the catalyst definition and

    Catalyst is a substance which accelerates the rate of a chemical reaction without undergoing any change in its chemical composition or mass during the reaction. The phenomenon of increasing the rate of a reaction with the help of a catalyst is known as catalysis.

  • Q : Solutions The normality of 10 lit.

    The normality of 10 lit. volume hydrogen peroxide is: (a) 0.176 (b) 3.52 (c) 1.78 (d) 0.88 (e)17.8

  • Q : Quastion of finding vapour pressure

    Vapour pressure of CCl425Degree C at is 143mm of Hg0.5gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl4 .Find the vapour pressure of the solution (Density of CCl4 = = 1.58g /cm2): (a)141.43mm (b)

  • Q : Some basic concepts of chemistry an

    an atom of an element is 10.1 times heavier than the mass of a carbon atom.What is its mass in amu?

  • Q : Preparation of normal solution Give me

    Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm

  • Q : Help 1) Chromium(III) hydroxide is

    1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer. 2) Explain how dissolving t

  • Q : Degree of dissociation The degree of

    The degree of dissociation of Ca(No3)2 in a dilute aqueous solution containing 14g of the salt per 200g of water 100oc is 70 percent. If the vapor pressure of water at 100oc is 760 cm. Calculate the vapor pr

  • Q : Determining concentration in ppm A 500

    A 500 gm tooth paste sample has 0.2g fluoride concentration. Determine the concentration of F in terms of ppm level: (a) 250 (b) 200 (c) 400 (d) 1000Answer: (c) F-ions in ppm = (0.2/500) x 106 = 400