# Identify the missing species and the radioactive decay process?

## a) $\text{_(94)^(239) "Pu" + ""_0^1 "n" -> ""_(Z)^(A) + ""_(36)^(94) "Kr" + 2_(0)^(1) "n}$ b) $\text{_(92)^(233) + ""_(0)^(1) "n" -> ""_(51)^(133) "Sb" + ""_(Z)^(A)? + 3_(0)^(1) "n}$ c) ""_(56)^(128) "Ba" + ""_(-1)^(0) "e" -> ""_(Z)^(A) ? + gamma d) $\text{_(Z)^(A) ? + ""_(-1)^(0) "e" -> ""_(35)^(81) "Br} + \gamma$

Jan 7, 2017

For convenience, the original questions are copied down below.

a) $\text{_(94)^(239) "Pu" + ""_0^1 "n" -> ""_(Z)^(A) + ""_(36)^(94) "Kr" + 2_(0)^(1) "n}$

When you set up the system of equations to solve this by asserting conservation of mass, just look across the top and across the bottom to get:

$239 + 1 = A + 94 + 2 \left(1\right)$
$94 + 0 = Z + 36 + 2 \left(0\right)$

Solving each one gives:

$A = 239 + 1 - 94 - 2 = 144$
$Z = 94 + 0 - 36 - 2 \left(0\right) = 58$

So, your isotope has an atomic number of $58$ and a mass number of $144$, meaning that it is cerium-144, $\textcolor{b l u e}{\text{_(58)^(144) "Ce}}$.

Nuclear fission is by definition absorbing a neutron and then splitting into two atoms and releasing a couple of neutrons.

b) $\text{_(92)^(233) + ""_(0)^(1) "n" -> ""_(51)^(133) "Sb" + ""_(Z)^(A)? + 3_(0)^(1) "n}$

Set up the system of equations:

$233 + 1 = 133 + A + 3 \left(1\right)$
$92 + 0 = 51 + Z + 3 \left(0\right)$

Solving these gives:

$A = 233 + 1 - 133 - 3 = 98$
$Z = 92 + 0 - 51 - 3 \left(0\right) = 41$

Atomic number 41 is niobium, so we have $\textcolor{b l u e}{\text{_(41)^(98) "Nb}}$. This is also nuclear fission for the same reason as in $a$.

c) ""_(56)^(128) "Ba" + ""_(-1)^(0) "e" -> ""_(Z)^(A) ? + gamma

The missing product is $\textcolor{b l u e}{\text{_(55)^(128) "Cs}}$.

Since the isotope absorbs an electron to combine with a proton and form a neutron, $A$ stays the same but $Z$ decreases by $1$, so I'd call this electron capture.

d) $\text{_(Z)^(A) ? + ""_(-1)^(0) "e" -> ""_(35)^(81) "Br} + \gamma$

This is just a variation on $l$. So, it's electron capture, and the missing reactant is $\text{_(36)^(81) "Kr}$.