PES - photoelectron spectroscopy measures the ionization energies of molecules when electron ejected from their different orbitals by absorption of a photon of known energy... This information uses to infer the energies of molecular orbitals.
The energy conserved when a photon ionized a sample sum of the ionization energy, I of the sample and kinetic energy of photoelectron, is the ejected electron.They have must be equal to the energy of the incident photon.. It's denotes hv.
hυ= 1/2meυ²+I equation.... (1)
In this equation which is one used for photoelectric effects equation
Ek=1/2meυ²= hυ-ɸ, written as,
hυ=1/2meυ²+ɸ.
It' can define in two ways. The first, photoelectrons may originate from one of a number of different orbitals series and different kinetic energies of the photoelectron will be obtained,
Each one satisfied hυ= 1/2 meυ²+Ii, where li, is the ionization energy of ejected electron from an orbital i, However by measuring the kinetic energies of the
Photoelectron, and knowing the frequency υ, these ionization energies can be determined. The Photoelectron spectra are interpreted in term called KOOPMAN'S THEOREM, which is states of the ionization energy with the energy of the approximation because it is ignore that fact that remains electrons adjust theirs distribution when ionization occurs.
This moleculecles are ionization energy several electronvolts even for valence electron, it's essential to work in at least the ultraviolet region of the spectrum and with wavelength of less than about 200nm. This radiation generated by discharge helium atom
The He atom (1s¹2p¹→1s²) it's lies at 58.43 nm. The corresponding to that photon energy is 21.22eV. It's use give rise to the technique of ultraviolet photoelectron spectroscopy (UPS) when core electrons are being studied photons of even higher energy are needed to expel them. X-rays are used this technique is denoted XPS.
This kinetic energy of the photoelectron measured using for electrostatic deflector that produces different deflections in the path of the photoelectrons as they pass between plates.
As they field of strength increases electron of different speeds and therefore kinetic energies reach the detector. The electron flux can be record and plotted against the kinetic energy to obtain the photoelectron spectrum
This is often observed that photoejection result in cation are excited vibrationally. But different energies are needed to excite different vibration states of the ion. The photoelectrons appearing with different kinetic energies. The resultants is vibrational fine structure, a progression of the lines with frequency space that corresponding to the vibrational frequency of the molecule Examples of vibrational fine structure in the photoelectron spectrum of HBr.
Photoelectrons ejected from N2 with He radiation have kinetic energies of 5.63V(1ev=8065.5 cm-¹, Helium radiation of wavelength 58.43nm has wave number 1.711×10⁵cm-¹ and therefore corresponds to an energy of 21.22eV. With Ii, in place of I, 21.22eV=5.63eV+Ii=15.59eV.This ionization energy is the energy needed to remove an electron from the occupied molecular orbital with highest energy of the N2 molecule, the 2σg bonding orbital.
KOOPMAN'S THEOREM :_
1) The basic assumption of this theorem is that molecular orbital appropriate for the parent molecule will be same as those the ionized molecule.
2) The vertical ionization energy for removed electron from a molecular orbital is equal to negative.
3)The stable orbital have negative eigen values.
4) Corresponding eigen value obtained from a (HFSCF) is calculation.
5) In the presence of zero field splitting the degeneracy is partially removed. because two transition are not degenerate and two peaks are observed in ESR spectrum.
(Eg) For Mn²+ ion: add number of unpaired electron :
1) For Mn²+ is a d⁵ system, it contains add number unpaired electron is kramer's degeneracy exist.
2) The zero field splitting produces three doubly degenerate
Spin state for Mn=5/2,n=1
3) The number spin states, =2nI+1
=2×1×5/2+=6
The spin state are +5/2,+3/2,-1/2,-3/2,-5/2.
4) Each of these spin states split into two singlets by the applied field producing size levels.
According to selection rule:
∆sm= ±1.
As a result of this splitting five transition are possibles:,
-5/2⇒-3/2
-3/2⇒-1/2
-1/2⇒+1/2
+1/2⇒+3/2
+3/2⇒+5/2
In a NQR experiment, radiation in the radio frequency region is employed to effect transition among the various nuclear energy levels.
No comments:
Post a Comment
Thanks for reading