2.04b Hydrogen-like atom of flourine

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En provenance du cours de University of Kentucky

Chemistry

615 notes

University of Kentucky

Chemistry

615 notes

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

À partir de la leçon

Week 2

Once we understand the electronic structure, we will be able determine how the periodic table can be used to compare the elements according to atomic size and various energies. We will conclude by seeing how these trends can be used to predict chemical properties of the elements.

2.01 Navigating the Periodic Table 12:32

2.02 Monoatomic ions and electron configuration 9:22

2.02a Electron Configuration of Ions 7:12

2.03 Effective Nuclear Charge and atomic and ionic radii 23:22

2.03a Isoelectronic and Radii 4:15

2.04 Ionization Energy 14:29

2.04a Wavelength of light emitted from first excited state 7:07

2.04b Hydrogen-like atom of flourine 7:59

2.05 Electron Affinity 7:45

2.06 Properties of Representative Elements 23:48

Rencontrer les enseignants

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

As you read through this problem it seems kind of complicated, but

I want to focus beginning with this information.

This is giving us an equation.

It tells us the amount of energy an electron has,

if it's a hydrogen-like atom.

Now, what do I mean hydrogen-like?

To be hydrogen-like it needs to have only one electron in it.

If it has only one electron in it, then we can use this equation.

And this is the, similar to an equation we did see in our lecture, but

it didn't have the Z term.

Z is a nuclear charge.

Now, if it's a hydrogen atom that's one.

That's why we didn't see it.

But let's look at this guy.

This is florine.

Florine has nine protons.

So it's Z is equal to 9.

So this could be how much energy as the number of protons in the nucleus increase.

The amount of energy an electron is going to have as

it is in its orbitals around that nucleus is going to increase as well.

Okay, so let us consider having an electron, and

this electron is sitting in the n equals 1.

And it's only one electron and it's in its ground state.

And let's determine the energy of that electron.

Okay?

And I want to put a little 1 here,

saying the electron is sitting in the n equals 1 state.

It is going to be negative Rydberg Constant,

2.18 times 10 to minus 18 joules.

Z is going to be 9 squared.

And, the electron is sitting in the one the first shell.

So that's a 1 and that's squared.

And when we multiply these values out.

We're going to get the value of a negative 1.765 times 10 to the negative 16 joules.

Oops, can't see that j.

Now what does that mean?

It's a negative number.

Well the way the notation goes is the electron has been just removed from

the atom.

But it, it does not got any kinetic energy associated with, with it.

It's just been pulled away and now it's no longer connected to the nucleus.

That would be an energy of zero.

And as it travels closer and

closer to the nucleus, it gets a, a more and more negative number.

Okay? So this electron is a negative value for

it, is a negative 1.765 times 10 to the minus 16 joules.

If I wanted to remove that electron from the nucleus, 'kay?

That would be called its first ionization energy.

Actually it wouldn't be

its first ionization energy because I've already removed eight electrons.

So this would be its ninth ionization energy.

Removing it from there.

I would have to supply exactly the amount of energy re, as that negative value is.

So that is the ionization energy to remove the ninth electron from a Fluorine atom.

So the negative number is associated with the electrons that are attached to or

connected with an atom.

And this is a value for the energy of the electron.

But to remove it would require you to supply this much ema,

energy in order to get it completely away from the nucleus.

'Kay. Now, we've got another one.

Let's consider another scenario.

So let's change colors for just a moment.

Let's pick hm, how about a purple.

Okay?

Lets move this electron now into the n equals 4 state.

Okay so we've promoted up, it is no longer sitting in the n equals 1.

We now have an n equals 4.

And we have an electron sitting right here.

If we wanted to know the energy of that electron, we'd call n,

with put a little 4 there.

And would a negative 2.18 times 10 to the minus 18 Joules.

This wouldn't change because that's the nuclear charge.

What would change is I would be dividing it by 4 squared, okay?

And when I do that, is again, it is a negative number.

But it's a negative 1.655.

No it's not, negative 1, when we multiply this out,

negative 1.104 times 10 to the minus 17 joules.

Now this is a less of a negative number.

It's not quite as big a negative number as this.

So it's sitting up here and it is a negative value, and

if I wanted to figure out how much energy it required to remove that electron,.

Okay, we would call it again it's the ninth electron but

it's coming out of an excited state.

So I would probably give it as 1.104 times 10 to the minus 17 joules.

It's a positive value so we move it.

So how much energy would it take to get it right disconnected from the atom.

So now in red I know how much energy it takes to remove this electron.

In purple, I know how much energy it takes to remove the electron from here.

Now let's read what it's asking.

What is the wavelength in nanometers of the emitted photon

in a transition from the n equals 4 to the n equals 1 state?

[NOISE].

Before we can determine the change in, or

the wave length we need to know the energy of the photon.

We need to know the change in energy of the electron.

Let's use again, another color here.

Let's, oops, what happened?

Hm, let's choose [SOUND]

on my screen.

We're going to promote an electron up to the n equals 4 state and

we're going to allow that to transition back down to the n equals 1 state.

When it does that, it's going to emit a photon of light, mm 'kay?

Now the energy of that photon of light will be

equal to the change in energy of the electron.

And that is going to be easily determined by subtracting red energy and

purple energy.

Okay?

So change in energy is going to be

1.104 times 10 to the minus 17 minus

1.765 times 10 to the minus 16.

And that's going to be equal to a negative 1.655, times 10 to the minus 16,

and that's going to be the change in energy of the electron.

We take the absolute value of that,

that's what’s going to give us the energy of the photon.

But what it wants to know is the wavelength of that light, of that photon.

So the wavelength equals, and if we look back at

previous problems that we've worked, I'll do it off to the side here.

We know that E equals hc over lambda.

So lambda equals hc over E.

We can plug all of our values in.

6.626 times 10 to the minus 34 joules times seconds,

times the speed of light, 3 times 10 to the 8 meters per second.

Divided by that energy that we figured out for the energy of the photon,

which is 1.655 times 10 to the minus 16 joules.

Now that'll give it to me in meters, but it really wants it in nanometers.

I'll go from meters to nanometers, 1 nanometer is 10 to the minus 9 meters and

this will give me a value of 1.20 nanometers.

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