0:02 the topic of this video is the bohr model
0:03 model
0:04 the learning objectives are on the
0:07 screen so before niels bohr
0:09 developed what we now refer to as the
0:11 bohr model uh
0:14 there was no application of quantized
0:15 energy states
0:18 to build the model of
0:20 of an atom such as the hydrogen atom
0:21 there is no
0:25 uh way to explain how there could be
0:27 discrete energy states there
0:30 so what the physicist kneels bohr
0:34 proposed is that the energy quantization
0:37 that we observe in different experiments
0:40 like the photoelectric effect
0:44 and also in black body radiation
0:46 was a result of discrete energy states
0:52 that are in orbit around a nucleus so
0:53 what niels bohr
0:57 derived independently of rydberg
1:00 is an equation that looks like this one over
1:00 over
1:02 lambda which is wavelength is equal to lowercase
1:03 lowercase
1:06 k this is a constant that i will give
1:08 the numerical value for in a second
1:10 that constant is divided by planck's
1:12 constant multiplied
1:14 by the speed of light another constant
1:16 and as i write this out it should
1:24 in that it has the overall
1:26 same form as the rydberg equation that i
1:29 previously discussed
1:32 the value for the lowercase k here this
1:33 particular constant
1:37 is 2.179
1:40 times 10 to the negative 18
1:44 joules okay so
1:47 uh the bohr model or niels bohr
1:49 in in developing the model came up with
1:50 this equation
1:53 using first principles and
1:56 what ended up helping get the
1:58 attention of other scientists at the
2:00 time was the fact that
2:04 niels bohr had um come up with a
2:06 an independent way of calculating what
2:07 was known as the
2:09 redbird constant and we still refer to
2:11 it as the rydberg constant
2:13 and the rydberg constant had been
2:15 experimentally measured
2:17 numerous times it was a very well known value
2:18 value
2:20 um and so the fact that niels bohr came
2:22 up with a way of using
2:25 um fundamental constants to just to to
2:26 calculate the
2:30 rydberg constant um was considered a
2:32 major success
2:33 and brought a lot of attention to the
2:35 bohr model so what does the bohr model uh
2:36 uh
2:39 look like essentially we have
2:42 a positively charged nucleus and there are
2:42 are
2:46 orbits around the nucleus
2:47 in this particular case of the hydrogen atom
2:49 atom
2:53 um maybe i'll draw one more orbit
2:55 these orbits as you get further and
2:57 further away from the nucleus
3:00 become higher and higher in energy so to
3:01 distinguish orbits from another
3:04 one another we give them numerical
3:05 integer values
3:08 one two and three the first
3:11 numerical value n equals one this is
3:12 what's referred to as the ground state
3:15 this is the lowest possible energy state of
3:15 of
3:19 an electron orbit now what can happen is
3:23 you can excite an electron from the
3:26 ground state up into an excited state
3:28 and when this happens it happens at a
3:30 very discrete energy
3:32 that that is the exact energy difference
3:34 between the n equals two
3:36 excited state and the n equals one
3:37 ground state so
3:41 anything above n equals one
3:48 that is these electrons in these uh
3:50 orbits that are further and further away
3:52 from the nucleus are at higher energies
3:53 they are excited relative to the ground state
3:55 state
3:57 okay so this can explain line spectra
3:59 and how might it explain
4:01 line spectra or specifically the line
4:07 um rather than just arbitrarily
4:09 assigning integer values like the
4:10 rydberg equation did it just you know
4:12 the rydberg equation just said okay well
4:13 if we just
4:16 use integer values not really knowing
4:18 what those integer values correspond to
4:20 we can recreate the the line spectrum of
4:21 the hydrogen atom
4:23 but what the bohr model says and what
4:25 bohr said is well those integer values
4:27 do have a meaning
4:30 they have a meaning of discrete
4:32 quantized energy states available to electrons
4:33 electrons
4:38 in orbit around a nucleus
4:39 so if we look over here in the case
4:43 where electrons move to higher energy as
4:44 light is absorbed this is the case where
4:46 we start with an electron down here at
4:49 the ground state n equals one
4:52 light comes in and excites an electron
4:53 to make and the electron can jump up
4:57 it can jump up to the n equals two state
4:57 or it could
4:59 skip the n equals two state if the
5:01 energy is high enough and it can go to
5:02 the three
5:05 four five in theory up infinitely
5:08 many states uh uh
5:11 but at some point the electron will escape
5:12 escape
5:15 the the the atom but that's a
5:17 we don't need to go there yet um it's
5:20 also important to note that you can also
5:23 have excitations uh from not just the
5:24 ground state but
5:26 in excited state to other excited states okay
5:27 okay
5:30 and these are very discreet energy jumps now
5:30 now
5:32 when we think about the line spectrum
5:34 that is light that is emitted
5:37 not light that was absorbed so in the
5:38 line spectrum of hydrogen what we see
5:40 are discrete energy bands and that
5:44 is because the um electrons have been
5:46 excited by electricity
5:47 uh electrical energy up to a higher
5:49 energy state
5:52 and then they can relax back down
5:55 so when the energy when the electron
5:58 is in an excited state and drops back
5:59 down to the ground state
6:01 light is emitted and that light is
6:03 emitted at an energy that corresponds to
6:05 the difference
6:07 between the excited state and whatever
6:10 state it relaxed to so you can have
6:13 a an electron go from any excited state
6:15 directly to the ground state
6:17 or you can also have these processes
6:18 where an excited state
6:21 will relax down to another excited state
6:23 and again this will release energy
6:26 that corresponds to the energy
6:29 associated with that transition
6:31 the other thing to point out here is an equation
6:32 equation
6:36 um that sort of uh captures this process
6:38 so if we want to ever consider
6:41 the the change in energy associated
6:43 between these transitions all we have to
6:44 do is
6:47 uh consider this delta e this is just
6:48 sort of a reworking
6:51 of the equation i already put up is
6:51 equal to
6:55 that k constant that i mentioned uh
6:58 multiplied by the this whole term which
6:58 is 1
7:01 over n 1 squared minus 1 over
7:06 n2 squared and this also by the way
7:08 since this is an energy value is equal
7:10 to hc
7:14 over lambda okay so let's go ahead and
7:15 put this uh
7:18 to to practice and i'm going to scroll
7:19 over to a practice problem
7:21 that i have here so go ahead and pause
7:23 the video now and you can write this out
7:25 in your notes if you need to
7:27 and i will go ahead and start solving it
7:28 okay so
7:30 what we need to do is find the energy in
7:32 joules and the wavelength in meters
7:36 of the line in the spectrum of hydrogen
7:37 that represents the movement of an
7:39 electron from the bohr orbit with n
7:41 equals four to the bohr orbit with n
7:42 equals six so we're going from
7:44 an excited state of n equals four to an
7:46 even higher excited state of n equals six
7:47 six
7:50 so um the uh since we're start our
7:51 starting point
7:54 is n equals four this will be our n1 and
7:56 our ending point
7:58 and two is is going up to that n equals 6
7:59 6
8:01 level so i'm going to just go ahead and
8:02 start by writing
8:04 let's calculate the the change in energy
8:06 for this jump from n equals 4 to n
8:07 equals 6.
8:10 so this is going to be equal to k times
8:13 1 over n 1 squared minus 1 over
8:16 n 2 squared so this is going to be equal
8:19 to 2.179
8:23 times 10 to the negative 18 joules
8:26 and this is uh n one i said it's going
8:28 to be four so that's four
8:31 squared minus um one over n two squared
8:33 so this is going to be six
8:37 squared so uh this will give us
8:40 2.179 times 10 to the negative
8:43 18 joules
8:46 multiplied by 1 over 16 minus 1 over
8:51 36 okay and if you do this and i
8:52 encourage you to practice
8:53 doing this in your calculator you should
8:57 get an energy value of seven point five
9:05 times ten to the negative twentieth
9:09 joules so this answer does make sense
9:11 just based off of the order of magnitude
9:13 of that scientific notation it's a very
9:15 small energy value
9:17 um so now the that's the first part of
9:19 the question uh the second part is
9:22 what is the wavelength in meters and in
9:23 what part of the electromagnetic
9:26 spectrum do we find this radiation
9:28 so let's go ahead and use the
9:30 relationship between
9:34 energy wavelength
9:36 planck's constant and the speed of light
9:37 so we know that
9:40 wavelength is equal to hc
9:46 so this is going to be equal to planck's
9:48 constant 6.626
9:51 times 10 to the negative 34 joule seconds
9:52 seconds
9:55 multiplied by the speed of light 2.998
9:57 times 10 to the 8th
10:00 meters per second all over this
10:03 energy value that we just calculated
10:04 five six six
10:06 times ten to the negative twenty joules
10:08 always do a unit check to make sure this
10:09 is cancel out joules we can sell
10:11 joules seconds will cancel out with
10:13 seconds we'll be left with units of
10:16 meters which is exactly what we want
10:21 and we should come up with a value of 2.626
10:22 2.626
10:26 times 10 to the negative 6 meters
10:29 that's our answer now if we look back to what
10:29 what
10:33 uh a figure that we uh that that shows
10:35 different parts of the electromagnetic
10:38 spectrum we'll see which region we find
10:40 this value in so we're at about 2.6
10:43 times 10 to the negative sixth meters
10:45 so this is a nice figure showing
10:46 different parts of the electro
10:50 electromagnetic spectrum we are at about
10:52 right here we are at about an order of
10:53 magnitude of 10 to the negative 6
10:56 meters for our wavelength so this puts us
10:57 us
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