--%>

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 : Problem on moles of solution The number

    The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75

  • Q : Question based on lowering of vapour

    Choose the right answer from following. The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be:  (a) 18.0 (b) 342 (c) 60 (d) 180

  • Q : Linde liquefaction process Liquefied

    Liquefied natural gas (LNG) is produced using a Linde liquefaction process from pure methane gas at 3 bar and 280 K (conditions at point 1 in figure below). A three-stage compressor with interceding is used to compress the methane to 100 bar (point 2). The first stage

  • Q : Explain physical properties of

    . Boiling pointsThe boiling points of monohalogen derivatives of benzene, which are all liquids, follow the orderIodo > Bromo > ChloroThe boiling points of isomeric dihalobe

  • Q : Ionization Potential Second ionization

    Second ionization potential of Li, Be and B is in the order (a)Li>Be>B (b)Li>B>Be (c)Be>Li>B (d)B>Be>Li

  • 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

  • Q : Molecular basis of third law. The

    The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be

  • Q : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0

  • 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