# 212 84Po→208 82Pb+AZX

What would the value of A be here?

212 84Po→208 82Pb+AZX

what would the value of z be here?

## Here is the way I write these. The atomic number goes on the left, the mass number on the right. So the Po would be

84Po212 ==> 82Pb208 + AXZ
where A is atomic number, Z is mass number, X is the element.

The left side must add up and the right side add up; i.e., atomic numbers must be the same on both sides and mass number must be the same.
On the left then we have 84 = 82 + A. A must be 2
On the right we have 212 = 208 + Z. Z must be 4
And X must be which element. Atomic number 2 must be He so that's an alpha particle that was ejected.

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## To determine the values of A and Z in the nuclear reaction: 212 84Po → 208 82Pb + AZX, we need to understand the conservation of mass number (A) and atomic number (Z) in a nuclear reaction.

In a nuclear reaction, the total mass number (A) and atomic number (Z) on both sides of the reaction should be equal.

From the given reaction:

212 84Po → 208 82Pb + AZX

We can see that the mass number (A) on the left side is 212, and on the right side, it is 208 + A (where A represents the mass number of the unknown particle).

Thus, the equation for the conservation of mass number can be written as:

212 = 208 + A

To solve for A, we can subtract 208 from both sides:

212 - 208 = A

A = 4

Therefore, the value of A in this reaction is 4.

For the value of Z:

From the given reaction:

212 84Po → 208 82Pb + AZX

We can see that the atomic number (Z) on the left side is 84, and on the right side, it is 82.

Therefore, the value of Z in this reaction is 82.

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## To determine the values of A and Z in the given nuclear equation, we need to understand a few key concepts:

1. Atomic number (Z): It represents the number of protons in an atom's nucleus, determining its element.

2. Mass number (A): It represents the total number of protons and neutrons in an atom's nucleus.

3. Isotopes: Atoms of the same element can have different numbers of neutrons, leading to different mass numbers. These variants are called isotopes.

Based on the given nuclear equation:

212 84Po → 208 82Pb + AZX

We can deduce that the 84Po atom (Polonium) on the left side of the equation undergoes radioactive decay (alpha decay) and transforms into a 82Pb atom (Lead) on the right side. Additionally, an unknown particle with a mass number of A and atomic number of Z is also produced.

By comparing the atomic numbers (Z) on both sides of the equation, we can determine the value of Z. In this case, Z remains the same, as both 84Po and 82Pb have the same atomic number (Z = 82).

However, the mass number (A) changes, indicating that there is a change in the number of neutrons. The difference in the mass number can be calculated as follows:

212 (mass of 84Po) - 208 (mass of 82Pb) = 4

Thus, the value of A in this nuclear equation is 4.

However, the value of Z (atomic number) for the unknown particle (AZX) is not provided in the equation, so we cannot determine its specific value.