Question

Write a nuclear equation to describe the neutron induced fission of Pu-239 to form Kr-89 and Ce-149 (this fission is induced by 2 neutron(s)). Determine how many neutrons are produced in the reaction.

Answer 1

Answer:

$\rm 2_{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, 3_{0}^{1}\text{n}; three$

Explanation:

If the fission is induced by two neutrons, the unbalanced equation is

$2\rm _{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, x_{0}^{1}\text{n}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation .  

Balancing the superscripts, we get

2 + 239 = 89 + 149 + x

      241 = 238 + x

          x = 3

The balanced nuclear equation is

$\rm 2_{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, \mathbf{3}_{0}^{1}\text{n}$

$\text{The fission produces } \boxed{\textbf{three}} \text{ neutrons}$

Answer 2

The number of neutrons produced in the reaction is 3.

Nuclear equation:

$2^1n_0+^{239}Pu---->^{89}Kr_{36}+^{149}Ce_{58}+3^1n_0$

In balancing nuclear equations- the sums of the superscripts and the subscripts must be the same on each side of the equation.  

On Balancing the superscripts, we get:

$2 + 239 = 89 + 149 + x\\\\241 = 238 + x\\\\x = 3$

The balanced nuclear equation is:

$2^1n_0+^{239}Pu---->^{89}Kr_{36}+^{149}Ce_{58}+3^1n_0$

Thus, number of neutrons produced in the reaction is 3.

Find more information about Nuclear equation here:

brainly.com/question/6199911