Rubidium is comprised of two isotopes, one of which has a natural abundance of 72% and contains 48 neutrons in the nucleus. given that the calculated atomic mass for rubidium is 85.5, how many neutrons are contained in the nucleus of an atom of the second isotope?

Since there are only two isotopes, then that means the other isotope has an abundance of: 100 - 72 = 28%. Let y be the mass of the first isotope, and x be the mass of the second. The equation would be

85.5 = 0.72y + 0.28x

Now, rubidium has 37 protons, and each proton has a mass of 1.00727647 amu. When neutral, it must also have 37 electrons in which each electron weighs 0.000548597 amu. Let n be the number of neutrons, in which each neutron weighs 1.008664 amu. The solution is as follows:

y = 37 (1.00727647) + 37 (0.000548597) + 48 (1.008664)

y = 85.7054 amu

Then, x would be:

85.5 = 0.72 (85.7054) + 0.28x

x = 84.972 amu

So,

x = 84.972 = 37 (1.00727647) + 37 (0.000548597) + n (1.008664)

Solving for n,

n = 47.27 ~ 47

Therefore, there are 47 neutrons in the second isotope.