Question

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.
$_{0}^{1}n+_{92}^{235}u
ightarrow4_{0}^{1}n + _{55}^{131}cs+_{an}^{mn}\text{symbol}
an=
mn=
symbol=
type = alpha$

Explanation:

Step1: Balance mass - number

The mass - number on the left - hand side of the nuclear reaction: For the neutron $\frac{1}{0}n$, the mass - number is 1, and for uranium $\frac{235}{92}U$, the mass - number is 235. So the total mass - number on the left is $1 + 235=236$.
On the right - hand side, for the four neutrons $4\frac{1}{0}n$, the total mass - number contribution is $4\times1 = 4$, and for cesium $\frac{144}{55}Cs$, the mass - number is 144. Let the mass - number of the missing nuclide be $AN$. Then $236=4 + 144+AN$, so $AN=236-(4 + 144)=88$.

Step2: Balance atomic - number

The atomic - number on the left - hand side: For the neutron $\frac{1}{0}n$, the atomic - number is 0, and for uranium $\frac{235}{92}U$, the atomic - number is 92. So the total atomic - number on the left is $0 + 92 = 92$.
On the right - hand side, for the four neutrons $4\frac{1}{0}n$, the atomic - number contribution is $4\times0 = 0$, and for cesium $\frac{144}{55}Cs$, the atomic - number is 55. Let the atomic - number of the missing nuclide be $MN$. Then $92=0 + 55+MN$, so $MN=92 - 55=37$.

Step3: Identify the element

The element with atomic - number $MN = 37$ is rubidium, and its symbol is $Rb$.
The type of reaction is not an alpha reaction. This is a fission reaction since a heavy nucleus (uranium - 235) is split into two lighter nuclei (cesium and rubidium) along with some neutrons.

Answer:

AN = 88
MN = 37
Symbol = $\frac{88}{37}Rb$
Type = Fission