Question

question 8
2 pts
in the reaction below, fill in the missing info for an, mn, & symbol for the missing piece. identify the type of nuclear reaction as well.
$\frac{13}{6}c+\frac{28}{13}al
ightarrow3\frac{1}{0}n+\frac{mn}{an}symbol$
an =
mn =
symbol =
type =

Explanation:

Step1: Calculate atomic number (AN)

In a nuclear - reaction, the sum of atomic numbers on the left - hand side equals the sum of atomic numbers on the right - hand side.
The atomic number of \(_{6}^{13}C\) is 6 and the atomic number of \(_{13}^{28}Al\) is 13. The atomic number of a neutron \(_{0}^{1}n\) is 0.
Let the atomic number of the unknown be \(AN\). Then \(6 + 13=3\times0+AN\), so \(AN = 19\).

Step2: Calculate mass number (MN)

The sum of mass numbers on the left - hand side equals the sum of mass numbers on the right - hand side.
The mass number of \(_{6}^{13}C\) is 13 and the mass number of \(_{13}^{28}Al\) is 28. The mass number of a neutron \(_{0}^{1}n\) is 1.
Let the mass number of the unknown be \(MN\). Then \(13 + 28=3\times1+MN\), so \(MN=13 + 28-3=38\).

Step3: Identify the element

The element with atomic number \(AN = 19\) is potassium (K).

Step4: Identify the type of nuclear reaction

This is a nuclear transmutation reaction as one or more nuclides are produced from the collisions of two atomic nuclei or a nucleus and a subatomic particle.

Answer:

AN = 19
MN = 38
Symbol = \(_{19}^{38}K\)
Type = Nuclear transmutation reaction