# 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.

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 $\text{Es}$ (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

$\boldsymbol{\text{_(Z)^(A_m) X = ""_(99)^(240) "Es}}$,

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

$\text{Atomic number = A + B + C} = 20 + 14 + 20 = 54 \Rightarrow 5 + 4 = 9$
$\Rightarrow 90 + 9 = 99$

$\text{Mass number = D + E + F} = 22 + 2 + 15 = 39 \Rightarrow 39 - 30 = 9$
$\Rightarrow 9 + 1 = 10 \Rightarrow 230 + 10 = 240$

So, we must devise a decay series for $\text{_99^240"Es}$.

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. b. Mark the box representing the starting nuclide

$\text{_99^240"Es}$ 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

$\text{_82^208"Pb}$ ($82 p , 126 n$).

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.

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

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.

$\text{_99^240"Es" → ""_82^208"Pb" + 8color(white)(l)""_2^4"He" + ""_1^0"e}$