identify the nuclide produced when americium-241 decays by alpha emission

from  a balanced decay equation we can identify the  nuclide formed

²⁴¹₉₅Am → ⁴₂He + ?

 te nuclide formed will have a drop of four from its mass number and a drop of 2 from its atomic number

²⁴¹₉₅Am → ⁴₂He + ²³⁷₉₃Np

 the nuclide formed is neptunium-237.