Design a decay series for a nuclide?

Take the sum of #A=20#, #B=14#, and #C=20#. Add the digits together, then add this result to #90# to get the atomic number of the element you are using. To get the mass number, add #D=22#, #E=02#, and #F=15#, then if it is over #30# subtract #30# until the result is below #30#. If the value is odd and not divisible by #4#, add #1#. Add this result to #230# to get the mass number.

2 Answers
Jul 25, 2017

Well, to find the atomic number, we add #A#, #B#, and #C# (whatever those are):

#20 + 14 + 20 = 54#

The sum of these digits is #9# (i.e. not two-digit, and is between #1# and #9#), so there is no need to repeat this. This number is added to #90# to give #Z = 99#, for #"Es"# (einsteinium).

To find the mass number, we add #D#, #E#, and #F# (whatever those are):

#22 + 02 + 15 = 39#

Since the result is over #30#, we subtract #30# until the result is #< 30#. We obtain #9#, then. Since the value is odd and not divisible by #4#, we add #1# to get #10#. Lastly, we add this value to #230# to get the mass number, #A_m = 240# (to distinguish it from #A = 20#...).

(If you don't follow, I am literally following the directions on the image.)

So, we currently have

#bb(""_(Z)^(A_m) X = ""_(99)^(240) "Es")#,

einsteinium-240. I'm guessing that's where you wanted me to stop.

Jul 26, 2017

Here's what I get.

Explanation:

Identify the starting nuclide

#"Atomic number = A + B + C" = 20 + 14 + 20 = 54 rArr 5 + 4 = 9#
# rArr 90 + 9 = 99#

#"Mass number = D + E + F" = 22 + 2 + 15 = 39 rArr 39 - 30 = 9#
# rArr 9 + 1 = 10 rArr 230 + 10 = 240#

So, we must devise a decay series for #""_99^240"Es"#.

1. Map a possible decay series

a. Highlight the isotopes that represent the Belt of Stability

Here is a chart I created in Excel.

It shows in yellow the isotopes of the elements from lead to einsteinium, as listed in ptable.org.

Isotopes

b. Mark the box representing the starting nuclide

#""_99^240"Es"# is the blue box at the upper right (#99 p, 141 n#).

c. Mark each isotope in your decay series

The blue boxes represent a decay series that includes as many as possible of the isotopes in the Belt of Stability and ends with

#""_82^208"Pb"# (#82 p, 126 n#).

d. Label the type of radiation emitted at each step

The particles emitted are

#α, β^"+", β^"+", β^"+", β^"+", α, α, α, β^"-", β^"-", β^"-", α, α, α, α#

2. Report the number of #α, β^"+"#, and #β^"-"# particles

The decay chain involves the emission of eight α particles, four #β^"+"# particles, and three #β^"-"# particles.

#""_99^240"Es" → ""_82^208"Pb" + 8color(white)(l)""_2^4"He" + 4color(white)(l)_1^0"e" + 3color(white)(l)_text(-1)^0"e"#

Note: I much prefer an alternate pathway involving the red blocks in the diagram.

It involves the emission of eight α particles and one #β^"+"# particle.

#""_99^240"Es" → ""_82^208"Pb" + 8color(white)(l)""_2^4"He" + ""_1^0"e"#