The umbrella of ozone in the upper atmosphere is formed from the photolysis of O2 molecules by solar radiation according to the following reaction:
$$o_2 + h\nu \rightarrow o + o $$
calculate the cutoff wavelength above which this reaction cannot occur (in nm) plzz give answer in nm only (nm).
Answers
Answer:
atmospheric chemistry, a null cycle is a catalytic cycle that simply interconverts chemical species without leading to net production or removal of any component.[1] In the stratosphere, null cycles and when the null cycles are broken are very important to the ozone layer.
One of the most important null cycles takes place in the stratosphere, with the photolysis of ozone by photons with wavelengths less than 330 nanometers. This photolysis produces a monatomic oxygen that then reacts with the diatomic oxygen producing ozone.[2] There is no net molecular or atomic change, however, this reaction converts the energy from the photons into heat energy warming the stratosphere.[3]
O3 + hv (λ < 330 nm) → O2 + O (1D)
O (1D) + M → O (3P) + M
O (3P) + O2 → O3
Net: hv → H
The null cycle can be broken in the presence of certain molecules, leading to a net increase or decrease in ozone in the stratosphere. One important example is NOx emissions into the stratosphere. The NOx reacts with both the atomic oxygen and ozone leading to a net decrease in ozone[2].This is particularly important at night when NO2 cannot photolyze.
NO + O3 → NO2 + O2
NO2 + O(1D) → NO + O2
Net: O3 + O(1D) → 2O2 (net loss of ozone)
Null cycles can also occur in the troposphere. One example is the null cycle that occurs during the day between NOx and ozone.
Tropospheric Null Cycle
O3 + NO → O2 + NO2
NO2 + hν → NO + O(3P)
O (3P) + O2 + M → O3 + M
Net: hv → H
This cycle links ozone to NOx in the troposphere during daytime. In equilibrium, described by the Leighton relationship, solar radiation and the NO2:NO ratio determine ozone abundance, maximizing around noon time.
Explanation:
Mark me as brainliest
Answer: 241 nm
Explanation: To determine the cutoff wavelength for the photolysis of O2 molecules, we need to know the bond dissociation energy of the O2 molecule. The bond dissociation energy for O2 is approximately 498 kJ/mol. We can use this value to calculate the minimum energy required for the reaction and determine the corresponding cutoff wavelength.
First, let's convert the bond dissociation energy to energy per molecule using Avogadro's number (6.022 x 10^23 molecules/mol):
(498 kJ/mol) / (6.022 x 10^23 molecules/mol) = 8.27 x 10^-19 J/molecule
Now we will use Planck's equation to find the cutoff wavelength:
E = h * c / λ
Where E is the energy per photon, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength.
Rearranging the equation to solve for λ:
λ = h * c / E
Plugging in the values:
λ = (6.626 x 10^-34 J s) * (2.998 x 10^8 m/s) / (8.27 x 10^-19 J)
λ ≈ 2.41 x 10^-7 m
Converting to nanometers (1 m = 1 x 10^9 nm):
λ ≈ 241 nm