--%>

Explain Photoelectron Spectroscopy.

The energies of both the outer and inner orbitals of atoms and molecules can be determined by photoelectron spectroscopy.

Energy changes of the outermost or highest energy electron of molecules were dealt with here in a different passion. The energies of other outer, valence shell electrons and even of inner electrons can also be determined. For such studies high-energy quanta of the far ultraviolet or of the x-ray region are required. Two techniques have been developed that are similar in principle but somewhat different in instrumentation and application.

One approach, known as photoelectron spectroscopy, uses the far-ultraviolet radiation from a helium discharge tube to expel an electron from the atom or molecule under study. This discharge tube provides an intense source of 58.4 nm radiation. Such radiation has quanta of energy 2046 kJ mol-1 or 21.21 eV. With such an ionizing beam, the electrons of the outermost electron shell of most atoms and molecules can be expelled. These electrons are most directly involved in the chemical, or bonding, characteristics of molecules. Thus photoelectron spectroscopy provides experimental results that are valuable adjuncts to the theoretical studies required.

A second experimental approach makes use of low energy or "soft", x rays. Various x-ray targets provide radiation with energies of several thousand electron volts. Such radiation can expel even the innermost electrons of atoms of molecules. Such electrons are bound with an energy that depends primarily on the nuclear charge of the atom to which they are held. The 1s electrons of the carbon atoms in any type of molecule, for example, have an ionization energy of about 1200 eV. However, this value depends somewhat on the chemical environment of the carbon atoms. Measurement of the ionization energies of such inner electrons thus reveals the various chemical environments of the atoms of a given type in the molecules of the sample under study. The results are therefore a guide to the nature of the molecule. As a result, this approach has many analytical applications. It is known as electron spectroscopy for chemical analysis (ESCA).

With either of the above two approaches, the sample is irradiated with high energy radiation, and electrons are expelled. Some of the energy of the incident radiation is carried off as kinetic energy of the expelled electrons. The energy required to produce a particular ionization is equal to the energy of the incident quanta less the kinetic energy of the expelled electrons. We can described this by

I = hv - εkin or εkin = hv - I, where I is one of the ionization energies and hv is the energy of the incident radiation. When high energy ultraviolet or x-rays ionizing radiation is used, atoms and molecules have a variety of ionization energies. Ionizations that require a large energy leave the expelled electrons with more kinetic energy. The pattern of electron kinetic energies thus yields a display of the various ionization energies of the atoms or molecules of the sample.

Orbital energies of atoms: a variety of ionization energies are found for the atoms of the sample.

Each ionization energy is a measure of the energy required to expel an electron from one of the orbitals of the atom. Or we can think of it as the energy released when an electron with zero kinetic energy is dropped into the empty orbital of the ion of the atom. Thus the ionization energies can be taken to be measures of orbital energies.

The ionization spectrum of argon illustrates the results obtained for free atoms. The ionizations can be described by the atomic symbols orbitals of the parent argon atoms from which an electron has expelled. The pattern of the orbital energies can be compared with the SCF values, notice that only the outermost orbital, 3p, has a low enough ionization energy to be studied by 58.4 nmhelium discharge radiation.

Orbital energies of molecules: now let us turn to the results obtained when the ionization energies of molecules are measured. We must be prepared for the ionized molecule. A, to be produced in the ground vibrational state or various excited vibrational states. These possibilities can again be treated in terms of the potential energy curves for the electronic states involved and the Frank-Condon principle.

Three types of situations are expected (1) if the ionized electron comes from a non bonding or an inner orbital, the potential energy curve for the molecular ion A+ should be little different from that for A. we expect v = 0 transitions to occur. A single ionization line would be expected. (2) If a bonding electron is removed, the bonding in A+ is expected to be weaker than that in A. the A+potential energy curve will be shifted in the direction of the longer equilibrium bonds. The weaker bonding will produce a more open potential energy curve. A relatively small vibrational spacing in the spectra band is expected. (3) If an electron of an antibonding orbital is expelled the opposite effect is expected. The nature of the ionization bonds produced for the ejection of non bonding, bonding, and antibonding electrons serves as a guide to the assignment of the bonds is resolved.

The ionization energy spectrums of N2 alongside the spectrum are the molecular orbital pattern and orbital designation, developed in the vibrational structure of the upper three bands is in keeping with the assignment. 

   Related Questions in Chemistry

  • Q : Question based on normality Provide

    Provide solution of this question. A 5 molar solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution : (a) 0.25 N (b) 1 N (c) 2 N (d) 7 N

  • Q : Net charge of a non-ionized atom

    Describe the net charge of a non-ionized atom?

  • Q : Problem based on normality Choose the

    Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The

  • Q : Molecular energies and speeds The

    The average translational kinetic energies and speeds of the molecules of a gas can be calculated.The result that the kinetic energy of 1 mol of the molecules of a gas is equal to 3/2 RT can be used to obtain numerical values for the

  • Q : Ddd 4) The addition of S2- ion to

    4) The addition of S2- ion to Fe(OH)2(s). Explain why the addition of S2- ion to Cr(OH)3(s) does not result in the formation of Cr2S3(s).

  • Q : Electrochemistry ( electrolysis of

    1. Define Faraday's first law of electrolysis 2. define Faraday's second law of electrolysis

  • Q : Question 6 A student was analyzing an

    A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu

  • Q : Colligative property related question

    Select the right answer of the question. Which of the following is not a colligative property : (a) Osmotic pressure (b) Elevation in B.P (c) Vapour pressure (d) Depression in freezing point

  • Q : What do you mean by the term dipole

    What do you mean by the term dipole moment? Briefly describe it.