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Current time:0:00Total duration:6:57

- [Voiceover] If you
didn't watch the last video because there was too much physics, I'll just quickly summarize
what we talked about. We went over the Bohr
model of the hydrogen atom, which has one proton in its nucleus, so here's the positive
charge in the nucleus, and a negatively charged
electron orbiting the nucleus. And even though this is not reality, the Bohr model is not
exactly what's happening, it is a useful model to think about. And so we just assumed the electron was going in this direction So counter-clockwise around
which gives our electron a velocity tangent to our circle, which we said was v in the last video. And in the last video, we
calculated this radius. So we calculated the
radius of this circle, and we said this was equal to r one. So r one turned out to be five
point three times 10 to the negative eleventh meters,
which is an important number. And we also derived this equation, right. So r for any integer n is
equal to n squared times r one, for example, if you wanted
to calculate r one again. So the first allowed
radius using the Bohr model is equal to one squared times r one. And so obviously one squared
is one so r one is equal to five point three times 10 to
the negative eleven meters. And so when n is equal to one, we said this was an electron
in the ground state, in the lowest energy state for hydrogen. We'll talk about energy
states in the next two videos. So this is a very important number here. So this is, this number right here, is the radius of the smallest
orbit in the Bohr model. In the previous video, we
also calculated the velocity or we came up with an
equation that you could use to calculate the velocity
of that electron. If you go back to the previous video, you'll see the equation that
we derived was the velocity is equal to the integer n
times Planck's constant divided by two pi m times r, and we figured this out
using Bohr's assumption for quantised angular momentum and the classical idea
of angular momentum. So if we plug in some numbers here, we can actually solve for
the velocity of this electron cause we're gonna take
this radius and we're gonna plug it in down here and then we know what these other numbers are. So we said n was equal to
one, so we're talking about n is equal to one so we're
going to plug a one to here. So this will be a one. The velocity is equal to
one times Planck's constant, six point six two six
times 10 to the negative 34 divided by two pi times m. And we're talking about
the electron so m was the mass of our electron, which
is nine point one one times 10 to the negative 31st kilograms. And finally, for n is
equal to one, this was our allowed radius so we can
plug this in for our radius, five point three times
10 to the negative 11. So if you do all that
math, I won't take the time to do it here, but you'll get a velocity approximately equal to,
approximate sign, two point two times 10 to the sixth and
your units should work out to meters per second
so that's the velocity. So going by the Bohr
model, you can calculate the radius of this circle
here so you can calculate this radius, and you can
also calculate the velocity. And,again, this isn't
reality but we'll use these numbers in later videos so
it's important to figure out where they came from. So this is the radius of
the smallest orbit allowed using the Bohr model but
you can have other radii, and we can calculate the
radii of larger orbits using this equation. So we're just gonna use
different values for n. So we started off with n is equal to one. Let's use the same equation and let's do n is equal to two. So let me go ahead and rewrite
that equation down here. Let's get some room. So r for any integer n is
equal to n squared times r one. Let's do n is equal to two here. So n is equal to two so let's
go ahead and plug in two. So we'd have two squared times r one. So r two, the second
allowed radii or the second allowed radius I should say,
is equal to four times r one. So if we're thinking about a picture, let's say this is the nucleus here and then this tiny, little
radius here is r one. If we wanted to sketch in
the second allowed one, it would be four times
that so I'm just going to approximate. Let's say that radius is four
times that so this is r two, which is equal to four times r one. And so we sketch in the radius
of this next radius here, this next allowed radius,
using the Bohr model. We could figure out mathematically
what that's equal to because we know r one is
equal to five point three times 10 to the negative 11 meters. And so if you do that calculation,
four times that number gives you approximately
two point one two times 10 to the negative 10th meters. So this is our second radius
when n is equal to two. Let's do one more when n is equal to three so let's get a little bit more room here. So when n is equal to three, this radius will be equal to
three squared times r one. So once again, we're just taking three and plugging it into here and so three squared is, of course, nine. So this would be equal to nine times r one so our next allowed radius
will be nine times r one. And I'm sure I won't get this accurate, but it's a lot bigger. So this will be r three is
equal to nine times r one. So I won't even attempt
to draw in that circle, but you get the idea. And we could do that math
as well, so nine times five point three times 10
to the negative 11 meters would give you approximately
four point seven seven times 10 to the negative 10th meters. And so these are the
different allowed radii using the Bohr model so you
can say that the orbit radii are quantised, only
certain radii are allowed so you couldn't get something in between. You couldn't have something
in between in here according to the Bohr model
so this is not possible. And these radii are associated
with different energies and that's gonna be really important and that's really why we're
doing these calculations. So we're getting these
different radii here and each one of these radii is associated with a different energy. So, again, more to come
in the next few videos about energy, which is
probably more important.