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AP®︎/College Chemistry
Course: AP®︎/College Chemistry > Unit 3
Lesson 12: Electronic transitions in spectroscopyWorked example: Calculating the maximum wavelength capable of ionization
If a photon has enough energy, it can completely remove an electron from an atom or molecule. In this video, we'll use the light equations (E = h𝜈 and c = 𝜆𝜈) to calculate the longest photon wavelength capable of removing an electron from a single atom of silver. Created by Sal Khan.
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Why does the question state "in the gas phase"? Would the energy be different in the solid?(2 votes)
Yes, the ionization energy would be different for the same substance but for a solid versus a gas. To be able to compare the ionization energy between different atoms, they are done in the gas phase to make it standard. The gas phase is chosen specifically because the attraction between the atoms is minimal when in the gas phase compared to other phases.
Hope that helps.(5 votes)
- At4:09Sal rewrites c/E/Planck's constant as c*Planck's constant/E. I don't understand how those two are equivalent(1 vote)
- Sal is dividing two fractions.
λ = c/(E/h) is the same as c/1 ÷ E/h
Dividing fraction means you multiply by the reciprocal of the second fraction.
c/1 x h/E = (ch)/E
Hope that helps.(3 votes)
Why does Sal round the energy? Wouldn't it be more precise to just straight up plug in the numbers into ch/((E/{Avogadro's Number})) => ch/v? In the problem sets, the solutions also tell us to round the energy first before plugging them in. Why is this?(1 vote)- You are correct in thinking that plugging everything as is given is more precise. I guess the sets are just there to make you understand the problems and not actually solve it in the most precise manner.(2 votes)
- At4:09when sal rewrites c/E/ Planck's constant as Planck's constant*c/E, shouldn't the E/ Planck's constant be in brackets to make it less confusing because it is it's own fraction as a denominator? For example c/(e/Plack's constant). I don't mean to be complacent I am just not very good at math and this is very confusing to me.(1 vote)
- Sal could have done so to be more clear, but we know from the previous substitution step that frequency, ν, is the same as energy divided by Planck’s constant, E/h. So it’s clear from the previous step that the dominator of the overall fraction is itself a fraction too. Sal also uses colors here to differentiate the numerator (cyan) from the denominator (pink).
Hope that helps.(1 vote)
4:09Video transcript
- [Instructor] We're told that
the first ionization energy of silver is 7.31 times 10
to the fifth joules per mole. What is the longest wavelength of light that is capable of
ionizing an atom of silver in the gas phase? All right, now before
I even ask you to pause and try to do this on your own, let's just remind ourselves
or try to understand what first ionization energy even is. This is the energy required
to get the highest energy or outermost electron
to escape from the atom. So it's not just going to
go to a higher energy level. It's just going to completely escape. You could view it as the
infinite energy level. And the reason why we're talking about the longest wavelength
of light is, remember, the longer the wavelength,
the lower the frequency, and the lower the energy. So this is saying really
what's the minimum frequency or the minimum energy that's associated with the longest wavelength of light for an atom of silver? So a couple things to pay attention to. They're giving us the
first ionization energy in terms of moles, not per atom. And then we just have
to remind ourselves all of our different ways of
connecting wavelength, frequency, and energy. Now, given all of this, I
encourage you to pause this video and see if you can figure this out. What is the longest wavelength of light that is capable of
ionizing an atom of silver in the gas phase? All right, now let's work
through this together. So the first thing to do is try to figure out the first
ionization energy per atom. And so maybe I'll write it like this. So the energy per atom, the first ionization
energy per atom is going to be equal to the ionization energy, 7.31 times 10 to the fifth joules per mole times what if we wanna
figure out per atom? This is per mole. Well, how many moles are there per atom? Well, we know that they're
this number of atoms per mole, so if we wanna know moles per atom, it's going to be one mole for every 6.022 times 10 to the 23rd atoms. I could write atoms here, and then that would
give us joules per atom, but we're just gonna get the
answer in terms of joules 'cause the moles are going to cancel out. And so this is going to
give us approximately... Let's see, we have three
significant figures here. 7.31 times 10 to the fifth divided by 6.022 times 10 to the 23rd power is equal to, and we have three significant figure here, so 1.21, approximately equal to 1.21 times 10 to the -18. 1.21 times 10 to the -18, and the units here are joules. This is joules per atom. So now how do we figure out wavelength? Well, as I alluded to, we might wanna use these equations here. We know that the speed of light is equal to the wavelength of that light times the frequency of the light. This is the lowercase Greek letter nu. This is not a V right over here. So if we wanna solve for wavelength, we just divide both sides by frequency. And so you get the wavelength is equal to the speed of light
divided by frequency. But how do you figure out
frequency from energy? Well, that's what this
top equation gives us. Energy is equal to Planck's
constant times frequency. So if you wanna solve for frequency, divide both sides by Planck's constant. So that top equation can
be rewritten as frequency, I'll write it here, frequency is equal to energy
divided by Planck's constant. And so we could take this
and substitute it over here, and we would get that
our wavelength is equal to the speed of light divided by energy divided by Planck's constant, or we can just rewrite this as being equal to the speed of light
times Planck's constant divided by, try to keep the colors
consistent, divided by energy. Well, we know what the speed of light is. It is 2.998, or it's approximately 2.998, times 10 to the eighth meters per second. We're gonna multiply that
time Planck's constant, which is 6.626 times 10 to the -34th joule seconds. And then we're going to divide that by the first ionization energy per atom, which we figured out right over here. So we're gonna divide that,
this E right over here. This is going to be, we figured it out, 1.21 times 10 to the -18 joules. Now let's make sure
all the units work out. So this seconds is going to
cancel out with this seconds. This joules is gonna cancel
out with this joules. And we're just gonna be left with meters, which makes sense. The wavelength can be measured in meters. And so let's just get our calculator out and calculate out what
this is going to be. 2.998 times 10 to the eighth times 6.626 times 10 to the -34. And then I'm gonna divide that by 1.21 times 10 to the -18. I think we deserve a
little bit of a drum roll. That gets us that. And let's see. If we have three significant
figures is our smallest amount in this calculation, so we're gonna go to these
three right over here. And so this is going to be 1.64 times 10 to the negative one, two, three, four, five, six, seven. So this is going to be approximately equal to 1.64 times 10 to the -7 meters. And we're done.