--%>

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 : Vapour pressure of methanol in water

    Give me answer of this question. An aqueous solution of methanol in water has vapour pressure: (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water (d) Less than that of water

  • Q : Question related to colligative

    The colligative properties of a solution depend on: (a) Nature of solute particles present in it (b) Nature of solvent used (c) Number of solute particles present in it (d) Number of moles of solvent only

  • Q : Group IV Cations Chromium(III)

    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.

  • Q : What is synthetic rubber and how it

    To meet human needs, scientists have started preparing synthetic rubbers. Besides having similar properties as natural rubbers they are tougher, more flexible and more durable than natural rubber. They are capable of getting stretched to twice its length. Though, it reverts to its original shape

  • Q : Determining mole fraction of water in

    A mixture has 18 g water and 414 g ethanol. What is the mole fraction of water in mixture (suppose ideal behaviour of mixture): (i) 0.1  (ii) 0.4  (iii) 0.7  (iv) 0.9 Choose the right answer from abo

  • Q : Molarity of HCl solution 20 ml of HCL

    20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is:  (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.

  • Q : Explain oxygen and its preparation.

    Karl Scheele, the Swedish chemist, was

  • Q : Normality of solution containing

    Can someone please help me in getting through this problem. Determine the normality of a solution having 4.9 gm H3PO4 dissolved in 500 ml water: (a) 0.3  (b) 1.0  (c) 3.0   (d) 0.1

  • Q : Strength of dilute acid of Sulfuric acid

    Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N

  • Q : Problem on decomposition reaction

    Nitrogen tetroxide (melting point: -11.2°C, normal boiling point 21.15°C) decomposes into nitrogen dioxide according to the following reaction: N2O4(g) ↔ 2 NO2(g)<