Question #3a474

1 Answer
Sep 7, 2017

Answer:

#""_39^88"Y"#

Explanation:

For starters, you know that you're dealing with a neutral atom, which means that the number of electrons that surround its nucleus must be equal to the number of protons present in its nucleus.

#color(blue)(ul(color(black)("no. of protons = no. of electrons"))) " " -> " neutral atom"#

This means that your atom has

#"no. of protons" = 39#

As you know, the number of protons located inside the nucleus of an atom, i.e. its atomic number, gives you the identity of the element, so grab a Periodic Table and look for the element that has an atomic number, #Z#, equal to #39#.

You should be able to identify this element as yttrium, #"Y"#.

So, you know that you're dealing with an isotope of yttrium. In order to be able to figure out the identity of this isotope, you must calculate its mass number, #A#.

#color(blue)(ul(color(black)("mass number = no. of protons + no. of neutrons")))#

In your case, you have

#A = 39 + 49 = 88#

To write this isotope in isotope notation, use

  • the name of the element
  • its atomic number
  • the mass number of the isotope

http://hydrogen.physik.uni-wuppertal.de/hyperphysics/hyperphysics/hbase/nuclear/nucnot.html

In your case, you have #"Y"# as the chemical symbol for the element, #Z = 39#, and #A = 88#, so you will end up with

#""_39^88"Y" " " -> " " "yttrium-88"#