[MUSIC]
Hi, and welcome to this in-depth extra
lecture on week one, your not going to
see my picture in this lecture, cause I
want you to watch what I write.
We're going to be going over five topics
in the next probably 20 minutes.
We're going to be converting among size
unit, estimating the numbers of things,
the surface
area of things, the size of things,and
finally
I'll talk about an atoms in a cluster.
There's a companion worksheet that you can
use.
That will provide some written material
that hopefully will
reinforce what we're going to be doing in
the next couple of slides.
Unlike our primary lectures, this resource
is really
provided to allow you to do well on the
in-depth quiz and also to see some problem
solving
if you have some questions about how to
tackle.
The quantitative part of the class.
Okay, so let's start with this first
example.
I'll let you go ahead and read that.
my advice is to go ahead and try to solve
this problem yourself, and then
watch how I do it.
Okay, so in this question you had to think
a little
bit about the geometry, because what
you're being asked to do.
Is to cut kind of like a bologna slicer,
for a piece of silicone
that's something like this in shape, so
your going to be cutting,
nanometer thick pieces of these AFM probes
and the probes
themselves are 35 nanometers.
In their thin width.
So, if you weren't sure how to do this
problem, odds
are it had something to do with getting
this geometry right.
But if we go ahead and say what this
problem is really asking us is how
many, so that can often be the more
complicated thing, is just to get to that
point.
So let me show you how we set this up.
We always write down what we're given,
we're given the centimeter.
And this is a conversion problem.
We know that one centimeter, is 10 to the
minus 2 meters.
We know that one meter is equal
to 10 to the positive 9 nanometers.
And now what we want to know
is.
35 nano meters is equal to 1 probe.
So we're assuming we have plenty of width
and length to deal
with and when we multiply all that out
what we're going to find.
Is 2.9 times 10 to the positive 7 times 10
to the minus 2.
Or on the order of 3 times 10 to the 5.
And that's
300,000 tips.
And a tip in this case is equal to a
probe.
Now what's also important to realize is a
step I did right here.
When you know that you have something
that's equal, like one
probe, you can rewrite that as one probe.
Over 35 nanometers.
And that is equal to one.
So it's part of a unit conversion.
It's just telling us if we want to go from
nanometers to probes, that's how we do it.
Now the other thing that I was doing that
I just want to point out explicitly.
So I was also cancelling units.
So if you're wondering how did I know to
put centimeters on
the bottom, and meters on the top, it was
a unit cancellation trick.
Now, let me go ahead and give you a
hint on this next example, which is on
your quiz.
In this example, it's just like what we're
doing
here, except now you have a four
centimeter post-it note.
And I'm asking you to figure out.
How many carbon nanotubes, if you laid
them flat down, would fit?
So it's very similar to the example that
we just did.
Okay, let's go on to the next slide.
So in this example, we're going to be
calculating finding the number of objects.
And the first thing that I want to
remind you is that this problem specifies
spherical particles.
So what that means is that the volume of a
single particle is going to be 4 3rds pi.
R cubed.
That's just a basic geometrical fact for r
spheres.
And how we're
going to do this is we're going to start
by calculating the volume of one particle
and I'm going to call that a nanoparticle,
so this is equal to np
for nanoparticle.
This is just going to be 4 3rds.
Pi, now R in this case remember, is 2.5
nanometers and I'm going to cube that, the
next thing I'm going to do is realize I'm
going to want all my units in centimeters.
So what we do is we write that conversion
but we cube it, because
this a nanometer cubed so that has to be
nanometer cubed to get centimeters cubed.
So when we multiply all of that out what
we're going to get is 6.5.
Times 10 to the minus 20.
And that makes sense because we only have
a single nanoparticle, so it should be
very small.
I also want to estimate that, sorry.
I also want to point out that what we're
doing here is we're really, estimating.
So I'm not going to be paying close
attention to
things like significant figures or where
the decimal point is.
That would be something I would do in a
formal class.
That was, for example, a whole semester,
but I'm just
trying to give you the basics of how to
approach this.
Okay, so that's the volume of a single
nano particle.
Now the next piece of information we have
then.
Is, we were given 10 microliters for our
injection
capacity, and that just comes from the
problem itself, there.
So we're going to shift, to centimeter
cubed.
Now what we're going to do is divide by
the volume of a single particle.
And here's how we write that.
I know the single particle is 6.5 times 10
to the minus 20.
Cenitmeter cubed and that is one nano
particle.
So when I multiply all of this out,
what I find.
And again these are estimates I'm not
carrying that 1.5 and I'll go ahead
and finish cancelling my unit so again
this term right here is an important
one because its saying the volume in a
single nano particle and this is
how you go from volume to the numbers of
particles is by figuring out.
The volume of a single nanoparticle, I get
a way to get rid of centimeter cubed, and
introduce nanoparticles as a unit.
Okay, this last one is going to be very
similar.
for this objects discussed, how many of
these particles would weigh one gram?
So in this case, you do need to know the
density.
So you're going to start with the one
gram.
So hint,
start with one gram,
take that to volume,
and then use the conversion above.
So you can use the same of one
nanoparticle is equal
to about 6 times 10 to the minus 20
centimeter cubed,
so your going to end up dividing just like
you did here,
but all of the work here will be a little
bit different.
Okay let's go on to the next problem.
So again I just want to make the point
that we're estimating.
Just so we don't have to be quite
as careful with all the little details,
and let's
think about this, so now we doing surface
areas,
so that's going to be a little bit
different.
We talked about that in lecture and so the
two pieces of data that I'm going to give
you.
Is that you can have a cube, and
then the surface area of the cube is,
there
are six spaces to the cube, and the area
of each face is 2 times R, where R is just
half the size of a face, squared.
If you have a sphere.
The surface area its just 4 pi R squared.
So in this problem
I ask you to consider both spheres and
cubes just to give you some practice.
Okay, so let's get started.
So in this problem, just like before, one
of our
first things we're going to do is,
calculate the volume
of the nanoparticle, because we're going
to need to know
how many nanoparticles that we actually
have in this entire situation.
So, funnily enough, if you use a cube,
then the volume of the cube, of
course is just going to be.
2R cubed.
And if you have a sphere, the volume of a
sphere is just 4 3rds pi r cubed.
So I'm going to go ahead and calculate
this out, just for kicks, as a cube.
So I'm going to start with 2 r cubed.
That's going to equal to 8, so that's
going to be 5 nanometers radius.
I could have just left that, but I didn't.
Remember I'm going to want everything in
centimeters cubed, so I already know.
That, I'm going to use this conversion.
And
when I work all that out, I get something
like this.
Now the other thing I'm going to have to
do is the surface area.
Of a nano-particle.
That's just going to be equal, assuming
also I have cube again.
To 6 times the edge length
squared.
So that's going to give me 2 times
2 is 4, times 6 is 24, times 5 nanometers.
Squared.
Now, I'm going to have to go from seven to
get to centimeters.
And that's just going to be squared.
So I have a slightly different conversion.
Well what I get for all of this is 6
times 10 to the minus 14 centimeters
squared.
So putting all that together then, I'm
going to take one gram.
I'm going to convert that to volume
through my density,
remember that density let's us go from
weight to volume.
And you can see the density number right
there, and that's exactly what I used
here.
I also know I'm writing gm sometimes,
really
to be accurate I should be writing g.
But I have a bad habit of writing gm, it's
the same as g, or grams.
Okay, so that gives me the volume.
Now
I'm going to go one particle.
So what I'm figuring out here in this part
of the calculation is, how many particles
are there?
So you always have to figure out how
many particles in these calculations to
start off with.
So now we have.
If we stopped, we'd have the number of
particles.
Okay, but what I care about is the surface
area of all of these.
So, I'm going to go back and
use this number and I'm going to say,
okay, that's how many particles I have.
And I know for every particle, I have this
much surface area, and when I put all that
together I get something on the order of,
100,000 centimeters squared.
Which is a whole lot of surface area
that's like 10 meters squared so that's a
lot.
Now I went ahead and did this for a
sphere which means I used this number and
this number.
And I did get something too different if I
used, if I did this for spheres.
I got something that look like to be about
60 meters squared.
So it's about an order of magnitude
difference.
If you take a sphere versus a cube.
Now on this last problem which is one of
your
quiz problems, it's really similar to what
I just did.
You're going to start.
By taking the 500 milligrams, you're
going to convert
that to volume using the density of gold.
Then you're going to do the same thing,
you're going to get number of particles,
and then you're going to get total surface
area.
And remember when you shrink, the particle
diameter.
From 1.15 to 1.5.
That's an order of magnitude.
So you should increase your surface about
an order of magnitude.
Okay.
Let's go to the next slide.
Okay.
So this is pretty simple but I was just
going to go over
it in case anyone had any questions based
on my size of things lectures.
So this is also very much related to the
nanoworld lecture.
As well
as the demo.
So when I did the demo, I sort of assumed
if I was 10 nanometers, how big would
things be?
So it's kind of a strange question, but
I'm about five foot six.
So 1.65 meters, in my world, and I was
writing kind of
the big world.
Is equal to one nanometer in the small.
So in this example, we're not doing 10
nanometers, we're doing
one nanometer.
So that's what's a little bit different
than the lectures.
And we're given a molecule, naphthalene,
and we're told it's five angstroms.
So I'm going to start with that.
So five angstroms, there's the symbol for
angstroms.
I knew that one angstrom is actually, I'm
going to
take a jump here, it's a tenth of a
nanometer.
And so it's an aromatic hydrocarbon, and
it's about 0.5 nanometers, roughly, in
this dimension.
Or five angstroms.
Okay so what I'm going to do is I'm going
to take this and I'm going to say this is
how big it is in the small world, in the
kind of world that it exists in so one
nanometers in the small world, so you go
to 1.56 meters.
In the big world.
Anyhow, if you work that out, what you can
figure out is that
naphthalene is going to be about, half
1.65 meters, or half my height.
So I need something that's about half my
height.
So I like a chair, about half my height or
kind of
more like the naphthalene because its a
plainer molecule, maybe a large kite.
So that's kind of how to think about it,
what you do is once your one nanometer
you sort of say is the object half my
size, twice my size, 10 times my size?
You can usually get there, but this is how
to do the math if you want to see it.
So this problem is pretty similar, you're
still one nanometer.
So I would take the three nanometers.
So whatever this is, it's
gotta be three times, taller than me.
Or you.
So you have to pick objects that are about
three times taller than you because
you are one nanometer so if it's three
times taller you got to think about it.
Okay lets go to the very last slide.
So this is actually some advanced work
that I just wanted
to give those students who were really,
really wanting to push themselves.
And what I wanted to do is actually
recognize that in a nanocrystal,
you saw that picture of Cadmium Selenide
with all the little dots?
They're actually atoms.
And so in a crystalite that's three
nanometers.
In diameter, there's going to be hundreds
of atoms inside of it.
So how do you calculate how many atoms are
held in a nanocrystal?
So, it's kind of interesting.
You have to realize,
that one of these atoms.
Which is, let's say in
this case, a gold atom.
A single atom weighs its atomic
number times the atomic mass unit which is
just a fixed
unit which is 1.6.
going to take 1.66
times 10 to the minus 24 grams
per kind of atom.
So, if you work this out, you can figure
out how much a gold atom weighs, and then
if you know how much the whole thing
weighs,
you can figure out how many gold atoms are
there.
So that's going to be our strategy.
So, let me get started.
So we know that the volume is equal to 4
3rds pi.
1.5 nanometers cubed times 10 to
the minus 21 centimeter cubed per
nanometer cubed.
All of that gives us 1.41 times 10 to the
minus 20 centimeter cubed.
I'm going to multiply that by the density.
And I'm going to get a weight.
And this weight is going to be the weight
of a gold particle.
And so then, I'm going to divide the
weight, of one nanoparticle.
By, 200 times 1.66 times 10 to the
minus 24 grams per atom.
So let me show you how this works out.
So four thirds pi R cubed, multiplied by
10 to
the minus 21.
Centimeters
cubed per nanometers cubed times 19.3
grams
per centimeters cubed.
Now we're onto weight of the entire
particle.
And I know that I have one gold atom
and it weighs 200 times 1.66 times 10 to
the minus 24 grams.
And going ahead and just doing my unit
analysis.
This is going to kill that, there's my
centimeters go away, grams go away, and
I'm going to get the number of gold atoms,
which is what I want.
And when I work all that out, I get
something on the
order of 820, which I like very much
because that's, I know,
about the right answer.
Now, my hint on this last question,
because this a question on your in-depth
quiz.
You're just going to do the reverse.
You're going to take 66 gold atoms.
You're going to figure out the weight.
You're going to figure out the volume.
You're going to get the radius.
You're going to have to take a cube root
to do that.
And then, you're going to double to get
the diameter.
So, have fun with that, and you know from
this
answer it better be a lot smaller than
that one.
I hope this has been useful for those of
you who are trying to do the in-depth.
I just wanted to do some of our
quantitative problem solving.
Please go ahead and try the in-depth quiz.
You get five tries on these questions.