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Worked example: Calculating the maximum wavelength capable of ionization | Khan Academy


4m read
·Nov 10, 2024

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 that as the infinite energy level.

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—an atom of silver?

So, a couple of 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. 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 want to figure out per atom? This is per mole. Well, how many moles are there per atom? Well, we know that there are 6.022 times 10 to the 23rd atoms in a mole.

So if we want to 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 going to get the answer in terms of joules because 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 figures here—so 1.21, approximately equal to 1.21 times 10 to the negative 18.

1.21 times 10 to the negative 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 want to 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 want to 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 want to 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 could just rewrite this as being equal to the speed of light times Planck's constant divided by—and trying 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 going to multiply that times Planck's constant, which is 6.626 times 10 to the negative 34 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 going to divide that—this e right over here—this is going to be, we figured it out, 1.21 times 10 to the negative 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 going to cancel out with this joules, and we're just going to 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 what this is going to be. 2.998 times 10 to the 8th times 6.626 times 10 to the negative 34, and then I'm gonna divide that by 1.21 times 10 to the negative 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 as our smallest amount in this calculation, so we're going to go to these three right over here. And so this is going to be 1.64 times 10 to the negative 1, 2, 3, 4, 5, 6, 7. So this is going to be approximately equal to 1.64 times 10 to the negative 7 meters, and we're done.

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