If radioactive neptunium decayed via the emissions of 1 alpha particle, 1 beta particle and 1 alpha particle, what would the daughter product be?
There's more than one isotope of Neptunium. The isotope that commonly undergoes alpha then beta then alpha decay is Np-237.
Neptunium has an atomic number of 93. Since a beta particle has two neutrons and two protons, Neptunium-237 would become Protactinium-233 upon emitting an alpha particle.
Protactinium-233 then emits a beta particle, which essentially came from a neutron turning into a proton and an electron. The emitted beta particle is the electron and the proton increases the atomic number by one, resulting in Uranium-233.
Uranium-233 then emits an alpha particle, becoming Thorium-229 (two less protons and 4 less in mass.)
To answer questions like this you need to make sure that the sum of atomic numbers and mass numbers is the same before and after for each decay step. For example, U-233 has an atomic number of 92 and mass of 233. The products, Th-229 and one alpha particle, have a total atomic number of (90+2)=4 and a total mass number of (229+4) =233.
Alpha emission reduces the atomic number by two and mass number by 4. Beta emission increases the atomic number by 1 and doesn't change the mass number.