Bohr Equation

Now the Boer equation is used to calculate the energy transition often electron as it moves from one shell toe another. So here we have two different bore equations here. We're gonna say in the first one this formula is used when dealing with two orbital levels, so they'll give you two and values and they're discussing energy here. We're gonna say changing energy of our electron equals negative are lower case E because we're dealing with energy times one over N squared final minus one over and squared. Initial here. Delta is the energy change for an electron in jewels. Here. We're going to say our he is our riper constant Here we're using e to differentiate it from the other are we're going to see in the other equation. So here this is our riper constant. Since it's in jewels, it's 2.178 times 10 to the negative jewels. Then we're gonna have our and final, which represents our final orbital level, and then we're gonna have an initial which represents our initial orbital or shell level. Now the next bore equation. This formula is used when dealing with again to orbital levels. Soul an initial and final still and you're dealing with wavelength here. We're gonna say one over wavelength equals negative are lambda just to show that we're dealing with wavelength here again, usually on your formula sheet and in your book they just used the variable are here We're changing it up slightly Just to show you that are sub e is when we're dealing with energy and our sub lambda is when we're dealing with wavelength and that's times one over and squared final minus one over and squared Initial So notice that these two formulas used the same portion here. What's changing is that we're dealing with energy here and wavelength here, and as a result of that, it has a change in my Rydberg constant value. Now since we're dealing with wavelength are Reiber constant will have units off meters in verse. In this case, are sub lambda equals 1.974 times 10 to the m in verse. So just remember, the first bore equation is when we're dealing with different shell numbers to shell numbers with energy. So remember the end values are your orbital levels or shell numbers, and we're dealing with this second bore equation when we have to orbital levels and wavelength and remember whether we're dealing with energy or wavelength. We're dealing with the Rydberg constant, but the values change based on if we're using jewels versus meters in verse.

here in states. What is the energy of a photon in jewels released during a transition from n equals four to n equals one state in the hydrogen atom. All right, so here, what we're gonna have is we're gonna say Delta E equals negative R E times one over N squared final, minus one over and squared Initial. We're using this form of the equation because we're dealing with energy. All we do now is we plug in the values that we know. First of all, the Rydberg constant, which is our when we're dealing with jewels. It's negative. 2.178 times 10 to the negative 18 jewels. Here we start off at N equals four. We're going down to n equals one. So our final and value is one so that be one squared and our initial would be 1/4 squared. So this comes out to be negative. 2.178 times 10 to the negative. 18 jewels. When we do 1/1 squared, minus 1/4 squared, that comes out 2.9375 When these two numbers multiply with each other, that's going to give us the energy involved with this electron transitioning transitioning from an n equals four to In n equals one state. So that comes out to be negative. 2.0 four 19 times 10 to the negative 18 jewels here, let's just do it in terms of 366 So negative 2.4 times 10 to the negative 18 jewels. In this question, all we have are shell numbers. But there are only one Sig Figs. So here, at the more comfortable, giving it three sig figs. Right, So here, negative 2.4 times 10 to the negative. 18 jewels are released as the electron moves from the fourth orbital level to the first orbital level.