Question #d4d86

1 Answer

Answer:

THe answer is d) #""_15^(32)"P"# and #""_16^(32)"S"#.

Explanation:

In order for two atoms that belong to different elements to be isobars, the must have the same mass number, #A#.

The mass number of an atom is equal to the sum of its protons and neutrons.

#A = Z + N#, where

#Z# - the number of protons the atom has, equal to the atomic number;
#N# - the number of neutrons it has;

For any element, the mass number is located in the upper left of the chemical symbol, while the atomic number is located in the bottom left of the symbol.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nucnot.html

So, all you have to do to determine which of those pairs is isobaric is to take a look at the number located in the upper left of the symbol.

For #""_6^(12)"C"# and #""_7^(14)"N"#, you have different mass numbers, #12# and #14#, respectively, so these two atoms are not isobaric.

The same can be said for #""_4^9"Be"# and #""_5^(11)"B"# (#9cancel("=")11#) and for #""_1^1"H"# and #""_1^2"D"# (#1cancel("=")2#).

However, if you take a look at the atoms in the last pair, you'll notice that their mass numbers are indeed equal, which makes them isobaric.

#""_15^(32)"P"# and #""_16^(32)"S"-> 32=32 => color(green)("isobaric")#