Answer
$^{134}Te$
Work Step by Step
The equation provided to us is
$^{233} \mathrm{U}+\mathrm{n} \rightarrow \mathrm{X}+\mathrm{Y}+b \mathrm{n}$
In this equation we are given that
Y = $^{100}$Zr and b = 2
$\therefore$ Our equation becomes
$^{233} \mathrm{U}+\mathrm{n} \rightarrow \mathrm{^{100}Zr}+\mathrm{Y}+ 2\mathrm{n}$
And now we have to find X
For our calculations,
We should remember that
For neutron Z = 0 & A = 1
For Uranium Z = 92
In this case, Zirconium has Z = 40
In such equations, Mass numbers & Z must be balanced to ensure that there is no mass loss on either side,
Therefore, Z for "X" will be 92-40-2 = 53
Tellurium has Z = 53
$\therefore$ X must be Tellurium
For A;
235+1-100 = 134
$\therefore$ X = $^{134}Te$
