--%>

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 : Molarity of Sodium hydroxide Select the

    Select the right answer of the question. Molarity of 4% NaOH solution is : (a) 0.1M (b) 0.5M (c) 0.01M (d) 0.05M

  • Q : Basic concepts Determination of correct

    Determination of correct mol. Mass from Roult's law is applicable to :

  • Q : Explain the polymers and its types.

    Polymers are the chief products of modern chemical industry which form the backbone of present society. Daily life without the discovery and varied applications of polymers would not have been easier and colourful. The materials made of polymers find multifarious uses and applications in all walk

  • Q : Explain Ionic Bond with examples. The

    The bonding in ionic molecules can be described with a coulombic attractive term.For some diatomic molecules we take quite a different approach from that used in preceding sections to describe the bonding. Ionic bonds are interpreted in terms of the coulom

  • Q : Explain Vapour Pressure Composition A

    A pressure composition diagram for a liquid vapor system can be used to show the composition of the liquid and equilibrium vapor.Vapor equilibrium data are useful in the study of distillations. It is of value to have diagrams showing not only the vapor pre

  • Q : Extensive property Choose the right

    Choose the right answer from following. Which one of the following is an extensive property: (a) Molar volume (b) Molarity (c) Number of moles (d) Mole fraction

  • Q : Question based on vapour pressure While

    While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than

  • Q : Strength of the Hydrochloric acid

    Provide solution of this question. 1.0 gm of pure calcium carbonate was found to need 50 ml of dilute HCL for complete reaction. The strength of the HCL solution is specified by : (a) 4 N (b) 2 N (c) 0.4 N (d) 0.2 N

  • Q : Organic structure of cetearyl alcohol

    Can we demonstration the organic structure of cetearyl alcohol and state me what organic family it is?

  • Q : Depression in the freezing point When

    When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o  (b) 0.80o