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# How much energy is required to ionize hydrogen (a) when it is in the ground state and (b) when it is in the state for which $n=3 ?$

## a) $$E_i = -13.6 \ eV$$b) $$E_i = -1.52 \ eV$$

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in this problem, we're gonna be looking at the ionization energies of hydrogen. Specifically, we're gonna be looking at the initial ah, and final states and the energy required to get between the estates. So we will have ah, hydrogen atom starting off with n equal, the one in the ground state. And then at the end, it will be ionized. So it'll be end going to infinity and the second is having an equal three. And then we also ionized this so it's gonna be and going to infinity. And so the question is gonna be how much energy does it take for each of these two processes? So we're talking about the change in states and the energy associated with it. You should probably think of this expression, which is Delta G is equal to h times F. Um, but since we're talking about ionization, we kind of have an expression that we use for this to compare states, Which is this the Rydberg expression with ah wavelength. So we want to replace this f with a lambda somehow which weaken do using this expression. So if we go ahead and do this will get each see divided by Lambda. And then from this point, you could rewrite Delta E by just plugging this in since it's already on the denominator for you. So you'll have a church. See, Time's the Rydberg constant, um, times the quantity one over ends of F squared minus one over ends of I squared. So this is the expression we can use for both of these scenarios that we're looking at here since, uh, those ends air just variables that we can plug in for. So let's keep this for reference appear as well as this. So these are the things we're gonna use. So for the first case, we're gonna have, um, let's call this Well, don't you want This is the 1st 1st case. So we're gonna have hh see ours of age. And so, like I said for ionization, we have to have and going to infinity, so we're gonna have one over infinity squared, which you might think is undefined. But that is effectively zero because it's one over a massive number. So it's gonna be effectively zero minus one over the initial state squared, which is one square one. So we're just going to get negative. H c times R h. So if you plug this in for Delta T one, you're going to get a value of negative 13.6 electron volts, but should make sense. And that should look like a familiar number two you because that's, um that's how the energy scales This is the base constant when you're talking about the scaling of energy with states. And it makes sense that if you're going from the ground state all the way up that that's one of your number. Okay, so for the second one for Delta Yay to we're gonna have a church. Well, actually, let's just go ahead and replace, uh, this constant from with 13.6, because we we know what it is now, right? Well, it would be positive. 13.6. So that's gonna be times, um, again, one over infinity squared. That's just gonna be zero minus 1/9. Because we have three squared. Right? So we're going. What we're gonna end up with is this 13 0.6 with the negative upfront from this term, I'm divided by nine. So overall, delta E two is going to be equal to negative, uh, negative 1.52 electron volts, which is one night are, um don't you want? Which should make sense, since energy does scale by and squared. So these are the ionization energies or hydrogen coming from the ground.

University of Michigan - Ann Arbor

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