Question

Identify the missing species and the radioactive decay process?

`a)` `""_(94)^(239) "Pu" + ""_0^1 "n" -> ""_(Z)^(A) + ""_(36)^(94) "Kr" + 2_(0)^(1) "n"`

`b)` `""_(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)` `""_(Z)^(A) ? + ""_(-1)^(0) "e" -> ""_(35)^(81) "Br" + gamma`

Answer 1

For convenience, the original questions are copied down below.

`a)` `""_(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(1)`
`94 + 0 = Z + 36 + 2(0)`

Solving each one gives:

`A = 239 + 1 - 94 - 2 = 144`
`Z = 94 + 0 - 36 - 2(0) = 58`

So, your isotope has an atomic number of `58` and a mass number of `144`, meaning that it is cerium-144, `color(blue)(""_(58)^(144) "Ce")`.

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

`b)` `""_(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(1)`
`92 + 0 = 51 + Z + 3(0)`

Solving these gives:

`A = 233 + 1 - 133 - 3 = 98`
`Z = 92 + 0 - 51 - 3(0) = 41`

Atomic number 41 is niobium, so we have `color(blue)(""_(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 `color(blue)(""_(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)` `""_(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 `""_(36)^(81) "Kr"`.