That will be 0.78*10 raise to the power of -1 nanometers
The wavelength of light : 4.14 x 10⁻⁵ m
<h3>Further explanation
</h3>
Radiation energy is absorbed by photons
The energy in one photon can be formulated as

Where
h = Planck's constant (6,626.10⁻³⁴ Js)
f = Frequency of electromagnetic waves
The energy of the light = 4.8 x 10⁻²¹ J
The wavelength of light :

The upper surface of the zone of saturation is called the water table
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
brainly.com/question/23038117
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The scheme is shown below, the steps involved are as follow,
Step one: Reduction: The carbonyl group of given compound on reduction using
Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation: The double bond present in starting compound when treated with
m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction: The epoxide is reduced to alcohol on treatment with
Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation: The hydroxyl group (alcohol) is
oxidized to carbonyl (ketonic group) using oxidizing agent
Chromic acid (H</span>₂CrO₄).