Consider the Bohr-model description of a hydrogen atom. (a)...

Consider the Bohr-model description of a hydrogen atom. (a) Calculate $E_{2}-E_{1}$ and $E_{10}-E_{9} .$ As $n$ increases, does the energy separation between adjacent energy levels increase, decrease, or stay the same? (b) Show that $E_{n+1}-E_{n}$ approaches $(27.2 \mathrm{eV}) / n^{3}$ as $n$ becomes large. (c) How does $r_{n+1}-r_{n}$ depend on $n ?$ Does the radial distance between adjacent orbits increase, decrease, or stay the same as $n$ increases?

Consider the Bohr-model description of a hydrogen atom.
(a) Calculate $E_{2}-E_{1}$ and $E_{10}-E_{9} .$ As $n$ increases, does the energy separation between adjacent energy levels increase, decrease, or stay the same? (b) Show that $E_{n+1}-E_{n}$ approaches $(27.2 \mathrm{eV}) / n^{3}$ as $n$ becomes large. (c) How does $r_{n+1}-r_{n}$ depend on $n ?$ Does the radial distance between adjacent orbits increase, decrease, or stay the same as $n$ increases?

Key Concepts

Asymptotic Behavior in Quantum Systems

As the principal quantum number increases, the properties of the system often approach a classical limit. Specifically, in the Bohr model, as n becomes large, the energy difference between adjacent orbits approaches a form that scales as 1/n^3. Analyzing such asymptotic behavior involves using approximations to simplify the description of the system in the high quantum number limit.

Radial Distance Between Orbits

In the Bohr model, the radius of an electron's orbit is proportional to n^2. Therefore, the difference between the radii of successive orbits increases with n, meaning that as n increases, the spatial separation between adjacent electron orbits becomes larger. This relationship highlights the spatial scaling in atomic structure as envisioned by the Bohr model.

Bohr Model of the Hydrogen Atom

This concept involves a model where electrons orbit the nucleus in specific, quantized orbits, each corresponding to a fixed energy level. The model postulates that electrons can only occupy certain orbits without radiating energy, and transitions between these orbits result in the emission or absorption of discrete energy quanta.

Energy Quantization and Level Differences

In the Bohr model, the energy of an electron is quantized and is inversely proportional to the square of the principal quantum number (n). The difference between adjacent energy levels, such as E(n+1) - E(n), decreases as n increases because of the n^2 dependency in the denominator, leading to an approximate dependence on 1/n^3 for large n. This explains why the energy separation between high-lying states is much smaller compared to that between lower-lying states.

Transcript

00:01 Problem we're talking about a bore model, a bore atom.

00:05 And our first order of business is to find our energy of e2 minus a we want.

00:15 So e2 minus e1 will equal negative 13 .60 electron volts per 2 squared minus negative 13 .60 electron volts over 1 squared.

00:32 And that will equal 10 .2, electron volts for e1 and e2.

00:38 For e 10 to e9 will have negative 13 .60 electron volts divided by 10 squared minus 13 minus negative 13 .60 electron volts per 9 squared.

00:59 And this will be 0 .0319 electron volts.

01:03 Of something else.

01:06 Does this energy separation, does the energy separation between adjacent? okay.

01:16 So the energy separation decreases as decreases for adjacent energy levels moving outward.

01:52 Okay.

01:54 And then for b, show that e .n plus y.

02:05 Minus n n approaches 27 .2 electron volts n cubed as n becomes large.

02:33 I would have read this, i wouldn't done it.

02:35 I don't like doing stuff like this.

02:36 Makes me confused.

02:40 Ok, so then how does show that this approaches? ok, so we know e.

03:10 Equals negative 13 .60 electron volts for n squared.

03:20 E n plus 1 minus e .n will equal negative 13 .60 electron volts over n plus 1 squared.

03:36 And this will equal negative 13 .6 or minus minus not minus minus negative 13 .6 over n squared.

03:54 So this will equal, i'm going to go to the next page to write this out.

04:00 This will equal 13 .60 times negative n squared plus n squared plus 2 n plus 1.

04:16 And in my denominator, i'm going to have n squared times n2 plus 2 n plus 1.

04:23 Then this will equal 13 .60 2n plus 1 over n4 plus 2n cubed plus n2...

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