### Video Transcript

94 40 Zr is a stable isotope of zirconium. What is the neutron-to-proton ratio of this isotope?

Isotopes are atoms that have the same number of protons but a different numbers of neutrons. The isotope we have been given is 94 40 zirconium. 40 is the atomic number, which is equivalent to the number of protons and, for a neutral species, also the number of electrons. 94 is the mass number for this isotope, which is the total number of protons and neutrons.

It’s important to note that if you were to find zirconium in the periodic table, the atomic number, or proton number, will remain the same. But instead of there being the mass number of a specific isotope, you will find the atomic mass, which is an average for all isotopes of that element based on their relative abundances. So we know that zirconium has 40 protons, and it has a total of 94 protons and neutrons. So, to find the number of neutrons in this isotope, we need to subtract 40 from 94, which is 54.

The question asks for the neutron-to-proton ratio. There are 54 neutrons and 40 protons. So the ratio is 54 to 40. We should simplify this by dividing both numbers by the smallest of these numbers, which is 40. If we divide 54 and 40 by 40, we get a ratio of 1.35 to one. So the answer to the question “What is the neutron-to-proton ratio of this isotope?” is 1.35 to one.

But let’s quickly check how feasible this answer is. The stability of a nucleus is primarily determined by its neutron-to-proton number ratio. The smallest atomic nuclei have a neutron-to-proton number ratio that is exactly or approximately equal to the number one. Medium-sized atomic nuclei tend to be stable when the ratio is approximately equal to 1.25. We are told in the question that the isotope is stable. We can tell from the periodic table that it’s roughly medium sized. And we have calculated that its neutron-to-proton ratio is 1.35, which is approximately equal to 1.25. So we can quite confidently say that the answer to the question “What is the neutron-to-proton ratio of this isotope?” is 1.35 to one.