Chemistry, asked by aadityajee2020, 10 months ago

1) for a given orbit r is directly proportional to 1/Z. staye for single electro species. 2) the distance between successive orbits became bigger and bigger as the value of n increses state for Hydrogen .state both conditions with equations NO SPAM​

Answers

Answered by urfriend55
0

The equation also shows us that as the electron’s energy increases (as n increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence on r in the Coulomb potential, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases, and it is held less tightly in the atom. Note that as n gets larger and the orbits get larger, their energies get closer to zero, and so the limits

imply that E = 0 corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state n = 1, the ionization energy would be:

With three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck’s constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr’s remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr’s model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics.

The figure includes a diagram representing the relative energy levels of the quantum numbers of the hydrogen atom. An upward pointing arrow at the left of the diagram is labeled, “E.” A grey shaded vertically-oriented rectangle is placed just right of the arrow. The rectangle height matches the arrow length. Colored horizontal line segments are placed inside the rectangle and labels are placed to the right of the box and arranged in a column with the heading, “Energy, n.” At the very base of the rectangle, a purple horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 2.18 times 10 superscript negative 18 J, 1.” At a level approximately three-quarters of the distance to the top of the rectangle, a blue horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 5.45 times 10 superscript negative 19 J, 2.” At a level approximately seven-eighths the distance from the base of the rectangle, a green horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 2.42 times 10 superscript negative 19 J, 3.” Just a short distance above this segment, an orange horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 1.36 times 10 superscript negative 19 J, 4.” Just above this segment, a red horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 8.72 times 10 superscript negative 20 J, 5.” Just a short distance above this segment, a brown horizontal line segment is drawn. A black line segment extends to the right to the label, “0.00 J, infinity.”

Figure 1. Quantum numbers and energy levels in a hydrogen atom. The more negative the calculated value, the lower the energy.

I think this notes will help you to find out the answer .

Similar questions