Question

I need help with this plz help

Answer 1

`\rm ^(103)_{\phantom{1}40}Zr`, zirconium-103.

Step-by-step 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* 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.