What did erwin schrodinger contribute to the atomic theory

Physics

2 answers:

lana [24]4 years ago

4 0

Answer:

Quantum or wave mechanical model

Explanation:

The great physicist Schrodinger Erwin developed the quantum mechanical model to improve on the existing Bohr's model that treated electrons as particles. He (Erwin) mathematically proved that the behavior of electrons in atoms is actually best treated as matter waves.

This model satisfactorily uses mathematical equations called wave functions to explain the behavior of the electrons.

One big advantage of this model is that while the Bohr's model uses one quantum number to describe how electrons are distributed within an atom which in turn determines the size of the orbit, it uses three quantum numbers (by putting the electrons in a 3-dimensional space) to describe the orbitals in which the electrons can be found.

These three quantum numbers used by Erwin's model are; principal, angular and magnetic quantum numbers. These numbers make it possible to describe the size, shape and orientation in space of the orbitals of electrons within atoms.

One big challenge with this model is that it gives the location where the electrons can be found and not where the electrons are exactly at.

Ivan4 years ago

3 0

Answer: QUANTUM MECHANICAL MODEL OF THE ATOM.

Explanation:Edwin Schrodinger is an Australian Physicist,who proposed the Quantum mechanical model of an Atom,this theory helped to give a better understanding of Bohr's model.

According to Bohr's model,it is believed that electrons are concentrated around the circular shell of the Nucleus of an Atom which Edwin Schrodinger improved by proofing that the Behaviors of electrons around an the Nucleus of an Atom can be explained by the Mathematical representation of their waves.

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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