Photochemistry process involve the initiation of a chemical reaction through the absorption of a photon by an atom or molecule.
When a molecule absorbs a photon of light, the energy is photon is transferred to the molecule. The energy of a photon is given by the Planck equation:
h = Planck constant = 6.626 × 10–34 J-s
c = speed of light in vacuum = 3 × 108 ms–1
v = frequency of light
and λ = wave length of light
So, S1, S2, T1 & T2 are electronic level.
S → singlet
T → triplet
Loss of excess electronic energy through the emission of a photon is known as rediative transition.
The process by which photons are emitted in radiat ive transit io n between S1 and S0 is known as fluorescence.
The process by which photons are emitted in radiat ive transit io n between T1 and S0 is known as phosphorescence.
The life time for phosphorescence is longer (10–6 s) than fluorescence (10–9 s) Photo physical reactions are corresponding rate expression
Quantum yield =
The Bear Lambert Law.
When a beam of monochromatic radiation of a suitable frequency passes through a solution, it is absorbed by the solution.
I0 = intensit y of incident light
I = intensity of transmitted light
and Ia = Intensity of the light absorbed = I0 – I
Absorbance of solution, A =
A = …(1)
where A = absorbance
C = concentration of solution
∈ = molar extinction coefficient or molar absorption coefficient (unit = concentration length–1)
l = path length
The absorbance of a solution is additive whereas the transmittance is multiplicative.
Problem. A monochromatic light is incident on solution of 0.05 molar concentration of an absorbing substance. The intensity of the radiation is reduced to one-fourth of the initial value after passing through 10 cm length of the solution. Calculate the molar extinction coefficient of the substance.
Sol. From bear Lambert law
2 × 0.3010 = 0.5 ∈
∈ = 1.204 dm3 mol–1 cm–1
Problem. A substance when dissolved in water at 10–3 M concentration absorbs 10 percent of an incident radiation in a path of 1 cm length. What should be the concentration of the solution in order to absorb go per cent of the same solution.
Sol. ∵ 10% absorbed then 90% transmitted
then T = = 90%
∵ 90% absorbed then 10% transmitted
C = 0.0218 mol dm–3.
Problem. In carbon-dating application of radio isotopes, 14C emits (1) β-particle (2) α-particle (3) γ--particle (4) positron
Correct answer is (1)
Problem. With increase in temperature, the Gibb’s free energy for the adsorption of a gas on a solid surface.
(1) Becomes more positive from a positive value
(2) Becomes more negative from a positive
(3) Becomes more positive from a negative value
(4) Becomes more negative from a negative value
Sol. From Langmuir isotherm, the fractional coverage is
∵ P ∝ T
Then higher the pressure or temperature lower the value of fractional coverage q.
R(g) + M(surface) RM (surface)
if q decreases then the format ion of RM decreases. i.e. rate of formation of RM decreases.
This indicate that Gibbs from energy of adsorption become positive.
So, increase the temperature, the Gibbs free energy of adsorption of a gas on a solid surface become more positive from a negative value.
The correct answer is (3).
Problem. One of the assumption made in the conventional activated complex theory is:
(1) Equilibrium is maintained between the reactants and the activated
(2) Equilibrium is maintained between the reactants and the product
(3) Equilibrium is maintained between the products and the activated complex
(4) Equilibrium is maintained between the reactants, the activated complex and the products
Sol. According to the Eyring the equilibrium is maintained between reactants and the activated complex
(AB)# is the activated complex.
The correct answer is (1)
Problem. For a reaction, the rate constant k at 27°C was found to be k = 5.4 × 1011 e–50 The activation energy of the reaction is
(1) 50 J mol–1 (2) 415 J mol–1 (3) 15000 J mol–1 (4) 125000 J mol–1
Sol. From Arrhenius equation
k = Ae-Ea / RT &(1)
k = 5.4 × 1011 e–50 …(2)
From equation (1) & (2) we get
The correct answer is (4)