Question

A curium-245 nucleus is hit with a neutron and changes as shown by the equation. Complete the equation by filling in the missing parts. 52 54 138 140 142 Xe Te Ce

Answer 1

The question is incomplete, the complete question is:

A curium-245 nucleus is hit with a neutron and changes as shown by the equation. Complete the equation by filling in the missing parts.

$_{96}^{245}\textrm{Cm}+_0^1\textrm{n}\rightarrow _{42}^{103}\textrm{Mo}+_{r}^{q}\textrm{p}+3_0^1\textrm{n}$

Answer: The complete equation formed is $_{96}^{245}\textrm{Cm}+_0^1\textrm{n}\rightarrow _{42}^{103}\textrm{Mo}+_{54}^{140}\textrm{Xe}+3_0^1\textrm{n}$

Explanation:

The general representation of an isotope is given as:

[rex]_Z^A\textrm{X}[/tex]

where,

A is the mass number, Z is the atomic number and X is the symbol of the element

In a nuclear reaction, total atomic number and total mass number on either side of the reaction remains the same

For the given chemical reaction:

$_{96}^{245}\textrm{Cm}+_0^1\textrm{n}\rightarrow _{42}^{103}\textrm{Mo}+_{r}^{q}\textrm{p}+3_0^1\textrm{n}$

On the reactant side:

Total atomic number = [96 + 0] = 96

Total mass number = [245 + 1] = 246

On the product side:

Total atomic number = [42 + r + 0] = 42 + r

Total mass number = [103 + q + 3] = 106 + q

Calculating for 'r' and 'q', we get:

$96= 42 + r\\\\r=96-42=54$

$246 = 106+q\\\\q=246-106=140$

Thus, the missing isotope is $_{54}^{140}\textrm{Xe}$

Hence, the complete equation formed is $_{96}^{245}\textrm{Cm}+_0^1\textrm{n}\rightarrow _{42}^{103}\textrm{Mo}+_{54}^{140}\textrm{Xe}+3_0^1\textrm{n}$