The following content is provided under a Creative
Commons license.
Your support will help MIT Open Courseware continue to
offer high-quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses, visit MIT Open Courseware at
ocw.mit.edu.
Hi I'm Sal.
Today we're going to solve problem number 6 of
exam 1 of fall 2009.
Now before you attempt a problem, there are a lot of
things that you should know, that is background material
that's going to help you solve the problem.
Especially during an exam, given that this was the
hardest on this exam.
So you should be able to finish this within
15 minutes or so.
So before you start the problem you should know what
the energy is of a charged particle in an electric field,
which is given by this equation.
The conversion from an electron volt to a joule,
which is just a conversion of energy and this should be
given to you on your table of constants.
Also the energy associated with an emission spectrum,
which is given by this equation where k is actually
the ionization energy of hydrogen, which is 13.6
electron volts and the z squared is the atomic number
of your atom of interest. And then the n f and the n sub i
are your transition states.
The DeBroglie wavelength, which is given by lambda, is
equal to a Planck's Constant divided by the momentum.
And always remember that you need to conserve energy.
This is what's going to get you the full
points on the problem.
So the problem, it's best to solve if you draw a little
image as you read the problem.
So the problem reads, atoms of ionized helium gas, He plus,
are struck by electrons in a gas discharge tube operating
with the potential difference between the
electrodes set at 8.8 volts.
So I'm going to go ahead and start drawing an image.
So let's say I have two plates where the voltage between
these two is set at 8.88 volts.
And what's happening here is that you're accelerating
electrons through here.
So you can imagine that--
say this is positively charges and this
is negatively charged--
that an electron starts from your negative plate and then
it accelerates toward your positive plate and you can
imagine your plate having hole or something small where your
electron can then exit and then be in free space under
the influence of no potential.
So then these electrons are actually being aimed towards
your helium plus ions.
So that's your target.
Now if I continue reading the problem, it says that the
emmision spectrum includes the line associated with the
transition from n equals 3 to n equals 2.
Calculate the minimum value of the DeBroglie wavelength of
scattered electrons that have collided with helium plus and
generated this line in the emission spectrum.
So it's an energy conservation problem.
And all this tells you is that, you have an electron
that is being accelerated through your potential.
The electron is going to strike a helium plus atom and
upon that you detect an emission spectrum.
Well what does that tell you?
That tells you that that electron actually excited an
electron in the atom.
The electron had to transition to a higher state to absorb
energy and then decay and in the process of decaying it
emits a photon, which is what you detect.
So this information is given to you.
But the problem asks to find the value of the DeBroglie
wavelength.
So since this an energy conservation problem, and this
is the very first step that happens and I know that my
energy equation for a charged particle is
simply q times the voltage.
So I know that q, which is a certain value of charge, the
product of these two when I'm multiplying by 8.8 volts, this
yields 8.88 electron volts.
So it's a new unit of energy where we just saw the
conversion.
And the reason why this takes this shape is because one
electron has the charge of 1.6 times 10 to the negative 19
coulombs and that's where the conversion from electrical
volts to joules come from.
So I'm going to go ahead and call this e sub i initial.
Because this is our initial energy that we're given.
So we just start with this energy, nothing else.
Nothing else was given from the problem.
So we want to go ahead and conserve this.
So our final state, the combination of whatever we are
solving, should not have a higher energy than what
initially we started with.
So because the problem says that we have an emission
spectrum, I'm going to go ahead and draw another image.
I'll label this one, initial and I'll
label this one, final.
So my final image, you can imagine a helium plus atom.
And the electron strikes the helium plus atom and we detect
a photon and you're getting a scattered electron.
That's what the problem says.
Scattered electron.
So what this tells me is that this energy now is going into
creating a photon and scattering an
electron from your atom.
So if I was to equate those two or what not I can already
say that, well in my final state I detect an emission or
an emission spectrum.
And I know what the transition says.
It goes from n equals 3 to n equals 2.
That's what the problem tells us.
So if I take my equation, my delta e of my emission
spectrum, is simply going to be negative k, which is 13.6
electron volts.
And I'm going to multiply by the atomic
number, squared now.
This is where people also make mistakes during the exam.
The atomic number is not 1.
It's actually 2 because we're talking about helium.
And the atomic number is not the number of electrons or
values in terms of an ion, but it's actually the number of
protons that you have. So helium has 2 protons.
So I can multiply this by 2.
And I'm going to square it.
And I know that I'm going from n and equals 3 to n equals 2.
So if I plug this into my equation I have 1 over 3
squared minus 1 over 2 squared, which is 9 and 4.
And if you plug this in, and assuming that your calculator
works and is correct, you get a value of 7.56 electron volts
because that was the unit that I was using.
Everything else is unitless, so my energy is now 7.56
electron volts for my transition.
Which covers this part for my photon.
So now we're left with what the energy of this is.
So what is the energy of the scattered electron?
Well if I equate the initial to the final I know that the
final is a composition of the transition
in my scatter state.
So I know that e initial, which is 8.88 electron volts,
this equals to the energy for the emission spectrum, which
is 7.56 electron volts.
So I'll put the units in square brackets so you don't
get confused.
And then plus the energy of the scattered electron.
So if I subtract the two, I get the energy of
my scattered electron.
And that pretty much gives me a value of--
after I subtract that--
e of scattered electron is equal to--
well the distance between these two is actually 1.32
electron volts.
But don't stop here because the problem didn't ask you to
figure out the energy of the scattered electron.
It actually told us to figure out what the DeBroglie
wavelength is of this scattered electron.
Now the fact that the electron is scattered, to me that means
that the only energy that this particle
has is kinetic energy.
So it's a classical form of energy.
So I can go ahead and I can equate this to just 1/2 mass
of the electrons times my velocity squared-- whatever
velocity it has.
Now I'd don't need to figure out what the velocity is.
This is just important to figure out what the DeBroglie
wavelength is.
So this is good.
Now if we look at what our DeBroglie wavelength is, we
know that lambda is equal to Planck's
Constant divided by p.
And Planck's Constant has a value of 6.6 times 10 to the
minus 34 with units of joules seconds.
So this is important too.
Always keep track of your units because dimensional
analysis will help you a lot when solving these problems. A
lot of the times you're given energy values in electron
volts but your solution will have a constant that doesn't
have an electron volt, that will have a joule, so you're
forced to make that conversion or else your problem is going
to be wrong.
OK so with that in mind, we know what h is.
But what about p?
Well p is momentum.
And classical momentum is just mass times your velocity.
So this is just mass times velocity, but this is the mass
times our velocity.
We know what the mass of the election is.
That's given on our table of constants.
But the velocity, we don't know what it is but we can
grab it from our classical energy.
So if I look at my energy--
from the side so I'm going to look at my energy--
I know that my energy--
I'm going to call this scat, for scattered--
the energy of my scattered electron is simply 1/2 the
mass of the electron times our velocity squared.
Now if I multiply both sides by 2m, I'm going to go ahead
and get us to the point where I can get an equation that is
just my mass times v.
So I'm going to multiply both sides by 2m, So I get 2 mass
of the electron, e scat of the electron, equals--
if I multiply this by 2m, the two cancel,
so I get an m squared.
m squared times v squared is essentially
just m times v squared.
And this becomes just mass of your electron v squared.
Now if I take the square root of both sides, I end up
getting a nice relation for just m times v.
So I know that m times v now is just simply equal to this--
canceled, the square root cancels the square.
The square root of 2 mass of the electron times the energy
of the scattered electron.
So if I go back to my DeBroglie
wavelength, I have m v.
So I can take that value as a function of energy and just
substitute it into my equation.
And that'll help us get the answer because we know what
the energy of the scattered electron is.
We calculated that on the first part.
So with that in mind we'll come over here.
Again this is 6.6 times 10 to the minus 34.
And the units are joules seconds.
And down here I have the square root of 2 times my mass
of the electron, which is on your table of constants.
And it's just 9.11 times 10 to the minus 31.
And this has units of kilograms. And the energy of
my scattered electron, which is 1.3 times 1.32 and this
units of electron volts.
Now this has units of electron volts.
This has units of joules.
This problem's a little bit--
now you're not going to get a good answer with that unless
you make the conversion.
I pointed it out, the bullet point in the beginning was
that 1 electron volt is equal to 1.6 times 10 to minus 19
joules, so if I simply multiply this by that
conversion, just 1.6 times 10 to the minus 19, this has
units now of joules per electron volt.
So this has units of joules per electron volt.
The equation turned out to be pretty long.
But this cancels the electron volt and now
has units of joules.
Which if you go through the math, you're going to go ahead
and at the end get a unit of just meters.
Because joule is just kilogram, meters squared per
second squared.
So all that factors out and you end up getting that.
If you go through the math and you get your math right, then
this should yield a value of 1.06 times 10
to the minus 9 meters.
So this is the value that will get you the right
answer on the exam.
Again this is lambda for your DeBroglie wavelength because
that's what the problem is.
So the problem asked for the DeBroglie wavelength but it
give you all this information to get to the point where you
needed to solve.
And by conserving energy and knowing the right conversion
factors between energies you're able to get an answer,
which is good.
It's really easy to complicate things and get a
wrong problem now.
I remember from grading the exams, the most common error
that people faced was actually letting the energy of the
electron be the energy of a photon.
Now that can't be because the energy of a photon is just
your Planck's Constant times a frequency.
And that's the energy for massless particles, your
electron has mass so if it's moving it's not
going to be h nu.
The energy is going to be kinetic energy.
1/2 mb squared.
If it's not in an electric field, that is.
So with that in mind just be confident when you solve these
problems and make sure that you know exactly what energy
equations to use to get the right answer.