Question

Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear reactions we must account for protons, neutrons, and electrons, as well as write out the symbols for various chemical elements. In a nuclear equation, the products and reactants are symbolized as AZX where X is the chemical symbol for the element, A is the mass number, and Z is the atomic number. There are two main rules to remember when balancing nuclear equations:

The total of the superscripts (mass numbers, A) in the reactants and products must be the same.
The total of the subscripts (atomic numbers, Z) in the reactants and products must be the same

What is the value of A in the following nuclear reaction?

237Np93 →233 Pa91 +AZX

Answer 1

The value of A in the given nuclear reaction 237Np93 →233 Pa91 +AZX is 237.

Step-by-step explanation:

In the given nuclear reaction, we are asked to find the value of A. In a nuclear equation, the products and reactants are symbolized as AZX where X is the chemical symbol for the element, A is the mass number, and Z is the atomic number. To balance the equation, we need to make sure that the total of the superscripts (mass numbers, A) in the reactants and products is the same, as well as the total of the subscripts (atomic numbers, Z). In this case, the mass number of the reactant 237Np93 is 237. To balance the equation, the mass number of the product AZX must also be 237. Therefore, the value of A in this nuclear reaction is 237.

Answer 2

The correct answer is - 4.

Step-by-step explanation:

As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -

The sum of the product mass number of products = mass of reactant

237Np93 →233 Pa91 +AZX is the equation,

Solution:

Mass of reactants = 237

Mass of products are - Pa =233 and A = ?

233 + A = 237

A = 237 - 233

A = 4

So the equation will be:

237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)