Electronic transitions are called vertical transitions. Comment.
See Answer →Raman rotational spectra of was observed with Hg light with Rayleigh line at
. Stokes lines were observed at
. Draw the rotational energy level diagram with Stokes and anti Stokes lines.
How would you prepare the following laboratory solutions?
i) 500 ml of 0.5 M Na₂CO₃
ii) 100 ml of 0.1 N of NaOH
See Answer →The frequencies of the first and second overtones respectively are exactly double and triple of the fundamental vibration frequency. Comment.
See Answer →Calculate the number of vibrational modes for a molecule of methane.
See Answer →Write a brief account note on Bioluminescence and its use in microscopy.
See Answer → has bond length
. Calculate the wavenumbers of the first four lines in its microwave spectrum.
[Given: mass of C=12.000 u and mass of N=14.007 u]
See Answer →What are Good Laboratory safety rules. Why they are important?
See Answer →Write the basic components and application of Fluorescence spectrophtometry.
See Answer →The rotational Raman spectrum of a diatomic molecule has the first line of a series of Stokes and anti-Stokes lines at , relative of the
of incident light. What would be its bond length.
Write the basic principle and application of NMR Spectroscopy.
See Answer →In Raman spectra, the Stokes line are more intense than the anti-Stokes lines. Why?
See Answer →What is density gradient centrifugation? Discuss briefly any of its method.
See Answer →Discuss the ionization of weak electrolytes with suitable example.
See Answer →Assuming that the force constants of H-Cl and D-Cl are the same, calculate the fundamental vibration frequency of D-Cl if that of H-Cl is .
Define the following:
i) Mole fraction
ii) Stocks shift
iii) Chromatogram
iv) Half life of radioisotopes
v) Resolving power
See Answer →What is meant by a selection rule? Give the selection rule for a rotating diatomic molecule modelled as a rigid rotator.
See Answer →What information about the molecule can be obtained from a rotation spectrum?
See Answer →A photoelectron spectrum of a compound containing nitrogen (N), aluminium (Al), and fluorine (F) shows three strong peaks at the following binding energies:
Peak X: ~686 eV
Peak Y: ~399 eV
Peak Z: ~73 eV
Find the best option from the choices given below that best explains the relative positioning of these peaks in terms of binding energy.
(i) Peak X corresponds to F 1s electrons, which have the highest binding energy because fluorine’s high electronegativity and small size give its 1s electrons a large effective nuclear attraction. Peak Y is N 1s and Peak Z is Al 2p.
(ii) Nitrogen has more valence electrons than fluorine or aluminium, therefore its peak appears at the highest binding energy.
(iii) The spectrum shows more aluminium atoms than nitrogen or fluorine, so its peak appears at the lowest binding energy due to dilution.
(iv) Aluminium has the highest nuclear charge, so its 2p peak should have the largest binding energy, but shielding effects reverse this trend.
Justify your answer.