Curium-243 as a alpha decay equation

An important thing we must keep note of is that an α-particle (alpha particle) is a helium nucleus. This means that it contains 2 protons and 2 neutrons; giving it a mass number of 4.

During alpha decay, an atomic nucleus emits an alpha particle.
Essentially, it decays into an atom with an atomic number of 2 less than its original amount and a mass number of 4 less than its original amount.

In our case, Curium-243 undergoing alpha radiation decay must abide the following equation:

²⁴³₉₆Cm → ⁴₂He + x

～ The (number at the top/mass number) of x must be
243 - 4 = 239

～ The (number at the bottom/atomic number) of x must be
96 - 2 = 94

～ Element 94 is Pu (Plutonium). Therefore, the equation is:
²⁴³₉₆ Cm → ⁴₂ He +  ²³⁹₉₄ Pu

Hope this helps! ッ

Anions are negative charged ions

60.0 - there are 3 significant numbers