Question

If 235 92 U goes through a nuclear chain reaction to produce one molecule of 93 36 Kr and one of 140 56 Ba, how many neutrons are released during this process?

Answer 1

The number of neutrons released is `=2`

Explanation:

The reaction is

`""_92^235U ->" _36^93Kr+_56^140Ba+x_0^1n`

There is conservation of the atomic number

`92=36+56+0xxx`

`92=92`

There is conservation of the mass number

`235=93+140+x`

`x=235-(93+140)`

`x=235-233=2`

The number of neutrons released is `=2`

Answer 2

Look at the ATOMIC notation....the subscripts and superscripts specify what is going on....

Explanation:

We gots `""_92^235U`...this SPECIFIES that in such an isotope there are 92 protons (note that this notation is a bit superfluous, because we specify that for `Z=92`, we gots the element uranium...)

The superscript is the mass number...i.e. the number of protons AND neutrons..., and this varies between isotopes. We know the former from `Z`...and so `"number of neutrons"=235-92=143` of the neutral, little blighters...

We finish with `""_36^93Kr` and `""_56^140Ba` and so we end with `"36+56=92 protons..."`, but with `"233 neutrons...."` and so in the nucular equation TWO neutrons have been lost...

And so we can complete the nuclear equation which almost balances mass and charge...

`""_92^235Urarr""_36^93Kr+""_56^140Ba+2n^0`