WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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Formula [27.3] tells us
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what the distance between slits and
the diffraction grating must be
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to have a maximum at a certain angle,
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given a particular wavelength of light
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and *m* is the order of the maximum
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so each wavelength will give many maxima
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and we want to find the first one
in part (a), in which case, *m* is 1.
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And we are told that the angle to this
first maximum is 20 degrees
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and so what is *d*, the distance
between the slits
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and we divide it by *sin Θ* on both sides
to solve for *d*.
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Now the wavelength is something
we have to figure out
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using equation [30.13]
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and we are asked in part (a) to use
the first wavelength
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in the Balmer series for hydrogen.
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So we have the reciprocal
of that wavelength
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is Rydberg's constant times 1 over
the final energy level squared
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minus 1 over the initial
energy level squared.
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And for the Balmer series,
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the final is 2
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and the initial has to be more than
the final in order for there to be a
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drop in energy level of the electron
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and therefore an emission of some
photon of some energy.
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So *n i* has to be more than *n f*
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and so if we want the first wavelength,
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we'll take the first number greater
than *n f* which is 3.
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So we plug in the number 3 here for *n i*.
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Now I solved for wavelength by
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taking the reciprocal of both sides
which means
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raise both sides to the exponent negative 1
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and this right-hand side, I just left it as
wrapped in brackets
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with an exponent negative 1 because
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you know, getting a common denominator here
and then flipping the fraction
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is more work than it's worth probably.
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So let's just do it like this.
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So we have the first wavelength in the
Balmer series is Rydberg's constant—
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1.097 times 10 to the 7 reciprocal meters—
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times 1 over the final
energy level squared
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minus 1 over the initial
energy level squared
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and all to the negative 1 giving us
656.34 nanometers.
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So that's the wavelength we plug in
to this formula here
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to get the distance between the slits;
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the order is 1—it's the first maximum—times
its wavelength divided by *sin* of 20 degrees
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which is 1.92 micrometers.
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Now in part (b), we are asked to figure out
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what will the angle be
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for the fourth line in the Balmer series?
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So *n i* has to be greater than *n f*;
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*n f* is definitely 2 because of the word
'Balmer', Balmer series
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and so *n i* can be 3, 4, 5, 6 and so on.
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And we want the fourth wavelength
so that means *n i* is 6;
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number 3 would be the first, 4, 5, 6,
6 is the fourth.
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So that's why we plug in
the number 6 there.
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And we rearranged this formula
to solve for *Θ* now
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so *Θ* is the inverse sign of the order times
the wavelength divided by
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the distance between the slits.
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And so we find out what the wavelength is
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by going Rydberg's constant times
1 over 2 squared minus
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1 over 6 squared and then
all to the negative 1
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which is 410.21 nanometers.
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And then that wavelength is
substituted in here
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and we are multiplying by 1 cause
it's the first order maximum
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for this fourth wavelength
in the Balmer series
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and dividing by the separation
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between the slits that we calculated
in part (a);
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that number's written here times
10 to the minus 6 instead of
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the prefix 'micro'
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and then we are taking the inverse sign
of all that to get the angle
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which is 12.3 degrees.
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And then part (c) says,
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at what angle will you find the second
order maximum for
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the first wavelength in the Balmer series?
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So that means *m* is 2
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and the wavelength is what
we calculated up here—
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first wavelength in the Balmer series—
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and then dividing by the separation
between the slits,
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taking the inverse sign of that
and we get 43.2 degrees.