If a radioactive nucleus alpha decays to yield a sodium24 nucleus in 14.8 hours, what was the identity of the original nucleus? Nuclear equation?

zXw ==> 11Na24 + 2He4

z = 11 + 2 = ? (13 and that is Al)
w = 24 + 4 = ?
So the equation is
13Al28 ==> 11Na24 + 2He4

To determine the identity of the original nucleus in an alpha decay process, we need to understand the concept of alpha decay and use the given information.

In an alpha decay, a radioactive nucleus emits an alpha particle, which consists of two protons and two neutrons. This emission results in a new nucleus being formed. In this scenario, we know that the newly formed nucleus is sodium-24 (Na-24).

To find the original nucleus that underwent the alpha decay, we can determine the atomic number and mass number of the original nucleus by subtracting the atomic number and mass number of the alpha particle from the respective values for sodium-24 (Na-24).

The atomic number of sodium is 11, so the original nucleus must have an atomic number 2 less than 11, which is 9. Thus, the original nucleus is fluorine, which has an atomic number of 9.

Additionally, the mass number of sodium-24 is 24, and the mass number of an alpha particle is 4. So the original nucleus must have a mass number 4 less than 24, which is 20. Therefore, the original nucleus is fluorine-20 (F-20).

The nuclear equation for this alpha decay process can be written as:

F-20 → Na-24 + α

This equation shows that an F-20 nucleus decays into a Na-24 nucleus by emitting an alpha particle (α).

In summary, the identity of the original nucleus is fluorine-20 (F-20), and the nuclear equation for the alpha decay process is F-20 → Na-24 + α.