A triply ionized beryllium ion, Be3+ (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (c) For the hydrogen atom, the wavelength of the photon emitted in the n = 2 to n = 1 transition is 122 nm (see Example 39.6). What is the wavelength of the photon emitted when a Be3+ ion undergoes this transition?

Video transcript

Welcome back, everyone. We are told that a hydrogen atom undergoes an electronic transition from the first excited state with a principle number of two to the second excited state with a principle number of three. And we are told when that happens, it absorbs electromagnetic radiation with a wavelength of 656 nanometers or 656 times 10 to the negative ninth meters. And we are tasked with finding in order for us to make the same electronic transition the wavelength of a singly ionized helium atom. So where do we start? Well, first, let's go ahead and observe our hydrogen atom scenario. We have that one over the la wavelength of our hydrogen atom is equal to the hydrogen rider constant times one over our uh first excited state, which is N equals two squared minus 1/3 squared. So what this gives us is one divided by 656 times 10 to the negative ninth is equal to our hydrogen ride. Beg constant times 1/2 squared minus 1/3 squared. Solving for our hydrogen ridr constant, we get one point oh 976 times 10 to the negative seventh meters to the negative first power changing colors. Here, we are going to use a very similar formula to find our wavelength of our singly ionized helium atom. This time we are going to have that one over hour wavelength is equal to the helium ride, beg constant times the same internal expression inside the parentheses is here. So we have 1/2 squared minus 1/3 squared. But what is our helium ride be constant? Well, our helium rider constant is simply going to be our atomic number squared times our hydrogen ride be constant. So we have two squared times one point oh 976 times 10 to the seventh, which gives us 4.39 oh times 10 to the seventh meters to the negative first power. With that in mind, let's go ahead and solve for our wavelength of our singly ionized helium ion. So what we have is 4.39 times 10 to the seventh times 1/2 squared minus 1/3 squared, which gives us that our wavelength of our helium ion is equal to nanometers, which corresponds to our final answer. Choice of B Thank you all so much for watching. I hope this video helped. We will see you all in the next one.