Answer:
[tex]\rm ^{103}_{\phantom{1}40}Zr[/tex], zirconium-103.
Explanation:
In a nuclear reaction, both the mass number and atomic number will conserve.
Let [tex]^{A}_{Z}\mathrm{X}[/tex] represent the unknown particle.
The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle [tex]^{A}_{Z}\mathrm{X}[/tex], [tex]A[/tex] is the mass number while [tex]Z[/tex] while [tex]Z[/tex] is the atomic number.
Sum of mass numbers on the left-hand side of the equation:
[tex]\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240[/tex].
Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:
[tex]\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137[/tex].
Mass number conserves. As a result,
[tex]A + 137 = 240[/tex].
Solve this equation for [tex]A[/tex]:
[tex]A = 103[/tex].
Among the five choices, the only particle with a mass number of 103 is [tex]\rm ^{103}_{\phantom{1}40}Zr[/tex]. Make sure that atomic number also conserves.
