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Hi I'm Sal.
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00:00:22,310 --> 00:00:24,980
Today we're going to solve
problem number 6 of
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00:00:24,980 --> 00:00:27,320
exam 1 of fall 2009.
11
00:00:27,320 --> 00:00:30,040
Now before you attempt a
problem, there are a lot of
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00:00:30,040 --> 00:00:33,680
things that you should know,
that is background material
13
00:00:33,680 --> 00:00:35,740
that's going to help you
solve the problem.
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00:00:35,740 --> 00:00:38,060
Especially during an exam,
given that this was the
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00:00:38,060 --> 00:00:39,560
hardest on this exam.
16
00:00:39,560 --> 00:00:41,540
So you should be able to
finish this within
17
00:00:41,540 --> 00:00:43,740
15 minutes or so.
18
00:00:43,740 --> 00:00:47,520
So before you start the problem
you should know what
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00:00:47,520 --> 00:00:49,800
the energy is of a charged
particle in an electric field,
20
00:00:49,800 --> 00:00:51,830
which is given by
this equation.
21
00:00:51,830 --> 00:00:55,640
The conversion from an electron
volt to a joule,
22
00:00:55,640 --> 00:00:58,810
which is just a conversion of
energy and this should be
23
00:00:58,810 --> 00:01:03,350
given to you on your
table of constants.
24
00:01:03,350 --> 00:01:06,970
Also the energy associated with
an emission spectrum,
25
00:01:06,970 --> 00:01:10,610
which is given by this equation
where k is actually
26
00:01:10,610 --> 00:01:13,740
the ionization energy of
hydrogen, which is 13.6
27
00:01:13,740 --> 00:01:18,140
electron volts and the z squared
is the atomic number
28
00:01:18,140 --> 00:01:22,320
of your atom of interest. And
then the n f and the n sub i
29
00:01:22,320 --> 00:01:24,340
are your transition states.
30
00:01:24,340 --> 00:01:27,360
The DeBroglie wavelength, which
is given by lambda, is
31
00:01:27,360 --> 00:01:31,370
equal to a Planck's Constant
divided by the momentum.
32
00:01:31,370 --> 00:01:33,430
And always remember that you
need to conserve energy.
33
00:01:33,430 --> 00:01:35,950
This is what's going
to get you the full
34
00:01:35,950 --> 00:01:37,400
points on the problem.
35
00:01:37,400 --> 00:01:41,940
So the problem, it's best to
solve if you draw a little
36
00:01:41,940 --> 00:01:44,690
image as you read the problem.
37
00:01:44,690 --> 00:01:49,170
So the problem reads, atoms of
ionized helium gas, He plus,
38
00:01:49,170 --> 00:01:52,350
are struck by electrons in a gas
discharge tube operating
39
00:01:52,350 --> 00:01:54,040
with the potential difference
between the
40
00:01:54,040 --> 00:01:57,060
electrodes set at 8.8 volts.
41
00:01:57,060 --> 00:01:59,580
So I'm going to go ahead and
start drawing an image.
42
00:02:05,910 --> 00:02:13,070
So let's say I have two plates
where the voltage between
43
00:02:13,070 --> 00:02:21,930
these two is set
at 8.88 volts.
44
00:02:21,930 --> 00:02:25,690
And what's happening here is
that you're accelerating
45
00:02:25,690 --> 00:02:26,750
electrons through here.
46
00:02:26,750 --> 00:02:29,660
So you can imagine that--
47
00:02:29,660 --> 00:02:30,992
say this is positively
charges and this
48
00:02:30,992 --> 00:02:32,570
is negatively charged--
49
00:02:32,570 --> 00:02:35,910
that an electron starts from
your negative plate and then
50
00:02:35,910 --> 00:02:39,750
it accelerates toward your
positive plate and you can
51
00:02:39,750 --> 00:02:42,940
imagine your plate having hole
or something small where your
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00:02:42,940 --> 00:02:46,550
electron can then exit and then
be in free space under
53
00:02:46,550 --> 00:02:50,890
the influence of no potential.
54
00:02:50,890 --> 00:02:57,330
So then these electrons are
actually being aimed towards
55
00:02:57,330 --> 00:02:59,350
your helium plus ions.
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00:02:59,350 --> 00:03:01,580
So that's your target.
57
00:03:01,580 --> 00:03:05,770
Now if I continue reading the
problem, it says that the
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00:03:05,770 --> 00:03:07,920
emmision spectrum includes the
line associated with the
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00:03:07,920 --> 00:03:11,310
transition from n equals
3 to n equals 2.
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00:03:11,310 --> 00:03:13,890
Calculate the minimum value of
the DeBroglie wavelength of
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00:03:13,890 --> 00:03:16,480
scattered electrons that have
collided with helium plus and
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00:03:16,480 --> 00:03:19,050
generated this line in the
emission spectrum.
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00:03:19,050 --> 00:03:21,950
So it's an energy conservation
problem.
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00:03:21,950 --> 00:03:26,010
And all this tells you is that,
you have an electron
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00:03:26,010 --> 00:03:28,780
that is being accelerated
through your potential.
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00:03:28,780 --> 00:03:32,600
The electron is going to strike
a helium plus atom and
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00:03:32,600 --> 00:03:35,960
upon that you detect an
emission spectrum.
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00:03:35,960 --> 00:03:36,910
Well what does that tell you?
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00:03:36,910 --> 00:03:41,360
That tells you that that
electron actually excited an
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00:03:41,360 --> 00:03:42,950
electron in the atom.
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00:03:42,950 --> 00:03:45,710
The electron had to transition
to a higher state to absorb
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00:03:45,710 --> 00:03:49,340
energy and then decay and in
the process of decaying it
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00:03:49,340 --> 00:03:52,800
emits a photon, which
is what you detect.
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00:03:52,800 --> 00:03:54,630
So this information
is given to you.
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00:03:54,630 --> 00:03:57,415
But the problem asks to find
the value of the DeBroglie
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00:03:57,415 --> 00:03:59,150
wavelength.
77
00:03:59,150 --> 00:04:03,220
So since this an energy
conservation problem, and this
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00:04:03,220 --> 00:04:07,510
is the very first step that
happens and I know that my
79
00:04:07,510 --> 00:04:09,530
energy equation for a
charged particle is
80
00:04:09,530 --> 00:04:13,220
simply q times the voltage.
81
00:04:13,220 --> 00:04:19,600
So I know that q, which is a
certain value of charge, the
82
00:04:19,600 --> 00:04:26,400
product of these two when I'm
multiplying by 8.8 volts, this
83
00:04:26,400 --> 00:04:29,730
yields 8.88 electron volts.
84
00:04:29,730 --> 00:04:32,120
So it's a new unit of energy
where we just saw the
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00:04:32,120 --> 00:04:33,140
conversion.
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00:04:33,140 --> 00:04:36,255
And the reason why this takes
this shape is because one
87
00:04:36,255 --> 00:04:40,880
electron has the charge of 1.6
times 10 to the negative 19
88
00:04:40,880 --> 00:04:42,835
coulombs and that's where the
conversion from electrical
89
00:04:42,835 --> 00:04:45,100
volts to joules come from.
90
00:04:45,100 --> 00:04:50,260
So I'm going to go ahead and
call this e sub i initial.
91
00:04:50,260 --> 00:04:53,740
Because this is our initial
energy that we're given.
92
00:04:53,740 --> 00:04:55,570
So we just start with this
energy, nothing else.
93
00:04:55,570 --> 00:04:57,370
Nothing else was given
from the problem.
94
00:04:57,370 --> 00:04:59,140
So we want to go ahead
and conserve this.
95
00:04:59,140 --> 00:05:03,690
So our final state, the
combination of whatever we are
96
00:05:03,690 --> 00:05:06,500
solving, should not have a
higher energy than what
97
00:05:06,500 --> 00:05:08,670
initially we started with.
98
00:05:08,670 --> 00:05:13,780
So because the problem says
that we have an emission
99
00:05:13,780 --> 00:05:19,370
spectrum, I'm going to go ahead
and draw another image.
100
00:05:19,370 --> 00:05:24,560
I'll label this one,
initial and I'll
101
00:05:24,560 --> 00:05:26,190
label this one, final.
102
00:05:29,660 --> 00:05:37,690
So my final image, you can
imagine a helium plus atom.
103
00:05:37,690 --> 00:05:44,990
And the electron strikes the
helium plus atom and we detect
104
00:05:44,990 --> 00:05:51,620
a photon and you're getting
a scattered electron.
105
00:05:51,620 --> 00:05:54,280
That's what the problem says.
106
00:05:54,280 --> 00:05:56,740
Scattered electron.
107
00:05:56,740 --> 00:06:03,050
So what this tells me is that
this energy now is going into
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00:06:03,050 --> 00:06:05,340
creating a photon and
scattering an
109
00:06:05,340 --> 00:06:07,510
electron from your atom.
110
00:06:07,510 --> 00:06:12,180
So if I was to equate those two
or what not I can already
111
00:06:12,180 --> 00:06:18,360
say that, well in my final state
I detect an emission or
112
00:06:18,360 --> 00:06:19,450
an emission spectrum.
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00:06:19,450 --> 00:06:21,670
And I know what the
transition says.
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00:06:21,670 --> 00:06:23,570
It goes from n equals
3 to n equals 2.
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00:06:23,570 --> 00:06:25,100
That's what the problem
tells us.
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00:06:25,100 --> 00:06:29,680
So if I take my equation, my
delta e of my emission
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00:06:29,680 --> 00:06:35,730
spectrum, is simply going to be
negative k, which is 13.6
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00:06:35,730 --> 00:06:36,980
electron volts.
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00:06:41,030 --> 00:06:43,400
And I'm going to multiply
by the atomic
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00:06:43,400 --> 00:06:44,670
number, squared now.
121
00:06:44,670 --> 00:06:47,400
This is where people also make
mistakes during the exam.
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00:06:47,400 --> 00:06:49,770
The atomic number is not 1.
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00:06:49,770 --> 00:06:52,000
It's actually 2 because we're
talking about helium.
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00:06:52,000 --> 00:06:58,200
And the atomic number is not
the number of electrons or
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00:06:58,200 --> 00:07:02,440
values in terms of an ion, but
it's actually the number of
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00:07:02,440 --> 00:07:05,140
protons that you have. So
helium has 2 protons.
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00:07:05,140 --> 00:07:07,630
So I can multiply this by 2.
128
00:07:07,630 --> 00:07:10,120
And I'm going to square it.
129
00:07:10,120 --> 00:07:16,160
And I know that I'm going from
n and equals 3 to n equals 2.
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00:07:16,160 --> 00:07:21,600
So if I plug this into my
equation I have 1 over 3
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00:07:21,600 --> 00:07:27,300
squared minus 1 over 2 squared,
which is 9 and 4.
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00:07:27,300 --> 00:07:31,530
And if you plug this in, and
assuming that your calculator
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00:07:31,530 --> 00:07:37,920
works and is correct, you get a
value of 7.56 electron volts
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00:07:37,920 --> 00:07:42,230
because that was the unit
that I was using.
135
00:07:42,230 --> 00:07:46,270
Everything else is unitless,
so my energy is now 7.56
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00:07:46,270 --> 00:07:48,050
electron volts for
my transition.
137
00:07:48,050 --> 00:07:50,800
Which covers this part
for my photon.
138
00:07:50,800 --> 00:07:53,900
So now we're left with what
the energy of this is.
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00:07:53,900 --> 00:07:56,920
So what is the energy of
the scattered electron?
140
00:07:56,920 --> 00:08:01,730
Well if I equate the initial to
the final I know that the
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00:08:01,730 --> 00:08:04,950
final is a composition
of the transition
142
00:08:04,950 --> 00:08:06,120
in my scatter state.
143
00:08:06,120 --> 00:08:16,110
So I know that e initial, which
is 8.88 electron volts,
144
00:08:16,110 --> 00:08:21,370
this equals to the energy for
the emission spectrum, which
145
00:08:21,370 --> 00:08:26,610
is 7.56 electron volts.
146
00:08:26,610 --> 00:08:30,270
So I'll put the units in square
brackets so you don't
147
00:08:30,270 --> 00:08:31,980
get confused.
148
00:08:31,980 --> 00:08:43,850
And then plus the energy of
the scattered electron.
149
00:08:43,850 --> 00:08:46,120
So if I subtract the two,
I get the energy of
150
00:08:46,120 --> 00:08:48,290
my scattered electron.
151
00:08:48,290 --> 00:08:51,870
And that pretty much gives
me a value of--
152
00:08:51,870 --> 00:08:54,080
after I subtract that--
153
00:08:54,080 --> 00:09:01,390
e of scattered electron
is equal to--
154
00:09:01,390 --> 00:09:06,090
well the distance between these
two is actually 1.32
155
00:09:06,090 --> 00:09:07,340
electron volts.
156
00:09:10,060 --> 00:09:12,330
But don't stop here because the
problem didn't ask you to
157
00:09:12,330 --> 00:09:14,690
figure out the energy of
the scattered electron.
158
00:09:14,690 --> 00:09:16,150
It actually told us to figure
out what the DeBroglie
159
00:09:16,150 --> 00:09:19,330
wavelength is of this
scattered electron.
160
00:09:19,330 --> 00:09:23,870
Now the fact that the electron
is scattered, to me that means
161
00:09:23,870 --> 00:09:26,780
that the only energy
that this particle
162
00:09:26,780 --> 00:09:28,140
has is kinetic energy.
163
00:09:28,140 --> 00:09:31,970
So it's a classical
form of energy.
164
00:09:31,970 --> 00:09:37,176
So I can go ahead and I can
equate this to just 1/2 mass
165
00:09:37,176 --> 00:09:40,880
of the electrons times my
velocity squared-- whatever
166
00:09:40,880 --> 00:09:42,010
velocity it has.
167
00:09:42,010 --> 00:09:45,160
Now I'd don't need to figure
out what the velocity is.
168
00:09:45,160 --> 00:09:47,960
This is just important to figure
out what the DeBroglie
169
00:09:47,960 --> 00:09:48,750
wavelength is.
170
00:09:48,750 --> 00:09:51,840
So this is good.
171
00:09:51,840 --> 00:09:56,050
Now if we look at what our
DeBroglie wavelength is, we
172
00:09:56,050 --> 00:10:02,460
know that lambda is
equal to Planck's
173
00:10:02,460 --> 00:10:06,640
Constant divided by p.
174
00:10:06,640 --> 00:10:15,140
And Planck's Constant has a
value of 6.6 times 10 to the
175
00:10:15,140 --> 00:10:21,450
minus 34 with units
of joules seconds.
176
00:10:21,450 --> 00:10:22,460
So this is important too.
177
00:10:22,460 --> 00:10:24,670
Always keep track of your units
because dimensional
178
00:10:24,670 --> 00:10:27,420
analysis will help you a lot
when solving these problems. A
179
00:10:27,420 --> 00:10:30,620
lot of the times you're given
energy values in electron
180
00:10:30,620 --> 00:10:35,050
volts but your solution will
have a constant that doesn't
181
00:10:35,050 --> 00:10:37,500
have an electron volt, that will
have a joule, so you're
182
00:10:37,500 --> 00:10:40,520
forced to make that conversion
or else your problem is going
183
00:10:40,520 --> 00:10:42,700
to be wrong.
184
00:10:42,700 --> 00:10:45,820
OK so with that in mind,
we know what h is.
185
00:10:45,820 --> 00:10:46,800
But what about p?
186
00:10:46,800 --> 00:10:47,980
Well p is momentum.
187
00:10:47,980 --> 00:10:51,520
And classical momentum is just
mass times your velocity.
188
00:10:51,520 --> 00:10:57,390
So this is just mass times
velocity, but this is the mass
189
00:10:57,390 --> 00:10:58,940
times our velocity.
190
00:10:58,940 --> 00:11:00,260
We know what the mass
of the election is.
191
00:11:00,260 --> 00:11:02,250
That's given on our table
of constants.
192
00:11:02,250 --> 00:11:04,970
But the velocity, we don't know
what it is but we can
193
00:11:04,970 --> 00:11:07,850
grab it from our classical
energy.
194
00:11:07,850 --> 00:11:11,250
So if I look at my energy--
195
00:11:11,250 --> 00:11:16,200
from the side so I'm going
to look at my energy--
196
00:11:16,200 --> 00:11:18,450
I know that my energy--
197
00:11:18,450 --> 00:11:21,984
I'm going to call this
scat, for scattered--
198
00:11:21,984 --> 00:11:27,810
the energy of my scattered
electron is simply 1/2 the
199
00:11:27,810 --> 00:11:30,980
mass of the electron times
our velocity squared.
200
00:11:30,980 --> 00:11:36,880
Now if I multiply both sides by
2m, I'm going to go ahead
201
00:11:36,880 --> 00:11:41,520
and get us to the point where I
can get an equation that is
202
00:11:41,520 --> 00:11:42,960
just my mass times v.
203
00:11:42,960 --> 00:11:47,380
So I'm going to multiply both
sides by 2m, So I get 2 mass
204
00:11:47,380 --> 00:11:55,360
of the electron, e scat of
the electron, equals--
205
00:11:55,360 --> 00:11:57,125
if I multiply this by
2m, the two cancel,
206
00:11:57,125 --> 00:11:58,820
so I get an m squared.
207
00:11:58,820 --> 00:12:00,790
m squared times v squared
is essentially
208
00:12:00,790 --> 00:12:03,250
just m times v squared.
209
00:12:03,250 --> 00:12:08,630
And this becomes just mass of
your electron v squared.
210
00:12:08,630 --> 00:12:16,580
Now if I take the square root
of both sides, I end up
211
00:12:16,580 --> 00:12:18,510
getting a nice relation
for just m times v.
212
00:12:18,510 --> 00:12:24,990
So I know that m times v now is
just simply equal to this--
213
00:12:24,990 --> 00:12:27,710
canceled, the square root
cancels the square.
214
00:12:27,710 --> 00:12:33,050
The square root of 2 mass of the
electron times the energy
215
00:12:33,050 --> 00:12:36,780
of the scattered electron.
216
00:12:36,780 --> 00:12:38,750
So if I go back to my DeBroglie
217
00:12:38,750 --> 00:12:40,550
wavelength, I have m v.
218
00:12:40,550 --> 00:12:43,830
So I can take that value as a
function of energy and just
219
00:12:43,830 --> 00:12:45,820
substitute it into
my equation.
220
00:12:45,820 --> 00:12:48,410
And that'll help us get the
answer because we know what
221
00:12:48,410 --> 00:12:49,900
the energy of the scattered
electron is.
222
00:12:49,900 --> 00:12:52,770
We calculated that on
the first part.
223
00:12:52,770 --> 00:12:56,380
So with that in mind we'll
come over here.
224
00:12:56,380 --> 00:13:01,590
Again this is 6.6 times
10 to the minus 34.
225
00:13:01,590 --> 00:13:05,760
And the units are
joules seconds.
226
00:13:05,760 --> 00:13:15,800
And down here I have the square
root of 2 times my mass
227
00:13:15,800 --> 00:13:18,480
of the electron, which is on
your table of constants.
228
00:13:18,480 --> 00:13:24,620
And it's just 9.11 times
10 to the minus 31.
229
00:13:24,620 --> 00:13:30,070
And this has units of kilograms.
And the energy of
230
00:13:30,070 --> 00:13:38,460
my scattered electron, which
is 1.3 times 1.32 and this
231
00:13:38,460 --> 00:13:41,640
units of electron volts.
232
00:13:41,640 --> 00:13:44,610
Now this has units of
electron volts.
233
00:13:44,610 --> 00:13:46,402
This has units of joules.
234
00:13:46,402 --> 00:13:48,010
This problem's a little bit--
235
00:13:48,010 --> 00:13:50,230
now you're not going to get a
good answer with that unless
236
00:13:50,230 --> 00:13:52,160
you make the conversion.
237
00:13:52,160 --> 00:13:54,880
I pointed it out, the bullet
point in the beginning was
238
00:13:54,880 --> 00:13:59,540
that 1 electron volt is equal
to 1.6 times 10 to minus 19
239
00:13:59,540 --> 00:14:03,300
joules, so if I simply
multiply this by that
240
00:14:03,300 --> 00:14:11,540
conversion, just 1.6 times 10
to the minus 19, this has
241
00:14:11,540 --> 00:14:18,800
units now of joules
per electron volt.
242
00:14:18,800 --> 00:14:23,740
So this has units of joules
per electron volt.
243
00:14:23,740 --> 00:14:25,220
The equation turned out
to be pretty long.
244
00:14:25,220 --> 00:14:27,780
But this cancels the electron
volt and now
245
00:14:27,780 --> 00:14:29,320
has units of joules.
246
00:14:29,320 --> 00:14:32,600
Which if you go through the
math, you're going to go ahead
247
00:14:32,600 --> 00:14:34,820
and at the end get a unit
of just meters.
248
00:14:34,820 --> 00:14:37,940
Because joule is just kilogram,
meters squared per
249
00:14:37,940 --> 00:14:39,380
second squared.
250
00:14:39,380 --> 00:14:44,260
So all that factors out and
you end up getting that.
251
00:14:44,260 --> 00:14:48,780
If you go through the math and
you get your math right, then
252
00:14:48,780 --> 00:14:54,610
this should yield a value
of 1.06 times 10
253
00:14:54,610 --> 00:14:59,920
to the minus 9 meters.
254
00:14:59,920 --> 00:15:06,550
So this is the value that
will get you the right
255
00:15:06,550 --> 00:15:07,660
answer on the exam.
256
00:15:07,660 --> 00:15:10,410
Again this is lambda for your
DeBroglie wavelength because
257
00:15:10,410 --> 00:15:11,740
that's what the problem is.
258
00:15:11,740 --> 00:15:15,120
So the problem asked for the
DeBroglie wavelength but it
259
00:15:15,120 --> 00:15:17,800
give you all this information to
get to the point where you
260
00:15:17,800 --> 00:15:18,890
needed to solve.
261
00:15:18,890 --> 00:15:25,050
And by conserving energy and
knowing the right conversion
262
00:15:25,050 --> 00:15:27,660
factors between energies you're
able to get an answer,
263
00:15:27,660 --> 00:15:29,740
which is good.
264
00:15:29,740 --> 00:15:34,940
It's really easy to complicate
things and get a
265
00:15:34,940 --> 00:15:36,450
wrong problem now.
266
00:15:36,450 --> 00:15:41,280
I remember from grading the
exams, the most common error
267
00:15:41,280 --> 00:15:45,410
that people faced was actually
letting the energy of the
268
00:15:45,410 --> 00:15:47,900
electron be the energy
of a photon.
269
00:15:47,900 --> 00:15:52,320
Now that can't be because the
energy of a photon is just
270
00:15:52,320 --> 00:15:54,950
your Planck's Constant
times a frequency.
271
00:15:54,950 --> 00:15:59,110
And that's the energy for
massless particles, your
272
00:15:59,110 --> 00:16:02,390
electron has mass so if
it's moving it's not
273
00:16:02,390 --> 00:16:03,410
going to be h nu.
274
00:16:03,410 --> 00:16:05,480
The energy is going to
be kinetic energy.
275
00:16:05,480 --> 00:16:06,370
1/2 mb squared.
276
00:16:06,370 --> 00:16:09,650
If it's not in an electric
field, that is.
277
00:16:09,650 --> 00:16:14,160
So with that in mind just be
confident when you solve these
278
00:16:14,160 --> 00:16:18,690
problems and make sure that you
know exactly what energy
279
00:16:18,690 --> 00:16:21,820
equations to use to get
the right answer.