Respuesta :
Here we have to predict the number of maxima will be there in the radial probability function of 4s orbital of the hydrogen atom.
There will be four (4) maxima in the radial probability function for the 4s orbital of the hydrogen atom.
We know that the probability of finding an electron at some point is proportional to the square of the Hamiltonian function (Ψ). The Ψ² is equivalent to the 4Πr²Rdr. Which is actually the probability that an electron will be found in this volume R²dV or 4Πr²Rdr. The r is the radius of the orbital.
Now when r = 0, R is not zero it has a maxima if we draw r vs radial probability curve for an electron. The r maxima of the curve is the 1st Bohr orbital for the hydrogen atom.
As the r increases the R approaches to zero exponentially and also 4Πr²Rdr, which relates to the size of the orbit. We observe as we move from 1s→2s→3s→4s, the maximum density of electron occurs at larger distance.
As for 4s the corresponding orbit is four so we will get four (4) maxima in the radial probability which will be apart from the nucleus of the hydrogen atom.
One can expect [tex]\boxed{{\text{four}}}[/tex] maxima in the radial probability function for the 4s orbital of the hydrogen atom.
Further Explanation:
Atomic Orbital:
The wave nature of electrons present in any atom is expressed by a mathematical function, known as atomic orbital. This wave function is used for determining the probability to find electrons in specific region around atomic nucleus.
Radial distribution function:
This provides the probability density to find electron in the region anywhere around the sphere surface that is situated at distance r from proton.
The formula to calculate area of sphere is as follows:
[tex]{\text{A}} = 4{{\pi }}{{\text{r}}^2}[/tex]
Where,
A is the area of sphere.
r is the radius of the sphere.
The probability to find an electron at any particular point or region is directly related to square of Hamiltonian function. This Hamiltonian function is represented by [tex]\Psi[/tex]. The value of [tex]{\Psi ^2}[/tex] is [tex]4{{\pi }}{{\text{r}}^2}{\text{Rdr}}[/tex], which in turn indicates the probability of finding electrons in the volume of [tex]{{4\pi }}{{\text{r}}^{\text{2}}}{\text{Rdr}}[/tex] or [tex]{{\text{R}}^{\text{2}}}{\text{dV}}[/tex], where r is taken to be the radius of orbital.
Since area of sphere is calculated by the expression of [tex]4{{\pi }}{{\text{r}}^2}[/tex], the radial distribution function becomes [tex]4{{\pi }}{{\text{r}}^2}{\text{Rdr}}[/tex].
When r is zero, R cannot be zero but it has maxima if the curve of r is drawn against radial probability. This maxima indicates the first orbital for the hydrogen atom. So four maxima are expected to be found in radial probability function for the 4s orbital in case of hydrogen atom.
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Atomic Structure
Keywords: atomic orbitals, area, sphere, zero, maxima, four, hydrogen atom, radial probability function, 4s orbital, radius, first orbital.