Question

I need help with this plz help

Answer 1

Answer:

$\rm ^{103}_{\phantom{1}40}Zr$, zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let $^{A}_{Z}\mathrm{X}$ 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 $^{A}_{Z}\mathrm{X}$, $A$ is the mass number while $Z$ while $Z$ is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

$\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240$.

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

$\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137$.

Mass number conserves. As a result,

$A + 137 = 240$.

Solve this equation for $A$:

$A = 103$.

Among the five choices, the only particle with a mass number of 103 is $\rm ^{103}_{\phantom{1}40}Zr$. Make sure that atomic number also conserves.