what nuclide is formed when 90sr undergoes β decay?

when STRONTIUM-90 under a beta decay,  there is a decreases in its number of neutrons and an increases in the number of protons, the atomic number increases by one, and the nuclide formed is yttrium-90.

The parent nuclide must have an atomic number less than the daughter nuclide by one while the mass number remain the same. in a beta decay the number of protons increases by one while the number of neutrons decreases by one, the nuclide formed is yttrium-90

 Since the total number of nucleons did not change the mass number also does not change.

 See the equation given

Strontium-90 decayed into  ytrrium-90 and electrons

⁹⁰₃₈Sr → ⁹⁰₃₉Yt + ⁰_-1e

  As you can see, the mass  number does not change  and hence the new atom formed is an isobar, an isobars are two or more atoms with same atomic number but different mass number.

 Let us balance the equation to see  if we are  wrong,

 Mass number is same on both side

 And atomic mass on the reactant side is 38 and on the product side is 39+ (-1)= 38.

Therefore the atomic number and mass number on the product an reactant side are equal.

 The equation is therefore balanced and we are not wrong.