A hydrogen atom is in a state with energy -1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Video transcript

Welcome back, everyone. We are making observations about the electron of a singly ionized helium ion. We are told that it is excited to a higher state by absorption of electromagnetic radiation. Now, the energy of our electron moving in the circular planar orbit is going to be negative 3.4 electron volts. And we are tasked with finding what is the electrons angular momentum? Well, we have a formula for this our formula states that our angular momentum is equal to the principle number times Plank's constant divided by two pi. But what is the number for our electron? Well, we are told that the energy of an electron at an excited state N for the singly ionized helium ion is going to be equal to negative 13.6 electron volts divided by our principle number squared times our nuclear charge squared where a single ionized helium ion, our nuclear charge is simply going to be two rearranging our equation. What we get is that our excited state or our principle number is going to be equal to the square root of negative 13.6 times Z squared all divided by E N. We have all of these values. So let's go ahead and calculate N here we have that N is equal to the square root of negative 13.6 times two squared divided by negative 3.4, which gives us a principle number of four. With that in mind, let me go ahead and change colors. Here. We can go ahead and calculate our angular momentum. We have that our angular momentum is equal to four times plank's constant, which is 6.63 times 10 to the negative 34th divided by two pi which when we plug into our calculator, we get 4.22 times 10 to the negative 34th kilograms meters squared divided by seconds corresponding to our final answer. Choice of e Thank you all so much for watching. I hope this video helped. We will see you all in the next one.