It takes about 10^16 years for just half the samarium-149 in nature to decay by alpha-particle emission. Write the decay equation, and find the isotope that is produced by the reaction.
62149Sm ==>24He + X
In a decay reaction the subscripts must add up and superscripts must add up. Therefore, the atomic number (the subscript) must be 60 for X and the superscript (the mass number) must be 145. Use that information to identify element X and write the symbol for it in place of X. I'll redo this if the subscripts and superscripts mess up.
To write the decay equation for the given scenario, we need to know the initial isotope before the decay. In this case, the initial isotope is samarium-149 (Sm-149). We also know that it decays by alpha-particle emission.
The decay equation can be represented as follows:
Sm-149 → 2α + ?
Where α represents an alpha particle (which is a helium-4 nucleus, He-4) and '?' represents the unknown isotope produced by the reaction.
Now, let's find the isotope produced by the reaction. The alpha particle consists of two protons and two neutrons, which means it has a mass of 4 atomic mass units (AMU).
The atomic mass of Sm-149 is approximately 149 AMU. When an alpha particle (4 AMU) is emitted, the resulting isotope should have an atomic mass of 145 AMU.
By referring to the periodic table, we find that an element with an atomic mass of 145 AMU corresponds to an isotope of europium (Eu), specifically Eu-145.
Therefore, the decay equation is:
Sm-149 → 2α + Eu-145