Answer:
[tex]_{40}^{103}Zr[/tex]
Explanation:
For any nuclear equation, we should utilize the law of mass conservation and the law of charge conservation. The sum of the masses on the left-hand side of the arrow should be equal to the sum of the masses on the right-hand side of the arrow (those are the superscripts for each nucleus). Similarly, the sums of charges should be equal (this is the law of charge conservation).
Let's say that the missing species is X with a mass of 'M' and charge of 'Z':
[tex]_{94}^{239}Pu+_0^1n\rightarrow _Z^MX+_{54}^{134}Xe+3_0^1n[/tex]
Find mass applying the mass balance law:
[tex]239+1=M+134+3\cdot1\\240 = M+137\\M=240 -137\\M=103[/tex]
This means our particle X has a mass of 103. Let's find the atomic number (the charge) same way:
[tex]94+0=Z+54+3\cdot0\\94=Z+54\\Z=94-54=40[/tex]
The atomic number of our nucleus is 40. That said, we have:
[tex]_{40}^{103}X[/tex]
Find the element in the periodic table with Z = 40. This is Zr. Meaning we can now identify it fully:
[tex]_{40}^{103}Zr[/tex]
