Identify the nuclide produced when neptunium-237 decays by alpha emission: 237 93np→42he + ?

Neptunium-237 has a Atomic mass of 237 g/mol. The reaction of decay is shown as,                        ²³⁷₉₃Np > ᵇₐX + ⁴₂He
where,                         X = Unknown Element
                                     ᵇ = Atomic Mass of Unknown Element
                                     ᵃ = Atomic number of Unknown Element
Solving for ᵇ,
                            237  =  ᵇ  +  4
                                 ᵇ  = 237 - 4
                                 ᵇ  = 233
Solving for ₐ ,
                            93  =  ₐ  +  2
                               ₐ  =  93 - 2
                               ₐ  =  91
Now, Putting values in ᵇ and ₐ of X,

                                                      ²³³₉₁X

Searching Periodic table for atomic number 91.

                                                       ²³³₉₁Pa

Result: The child Isotope is Protactinium (Pa) having atomic number 91 and Atomic mass 233 g/mol

The nucleus inside an atom consists of positively-charged  protons  and neutrally-charged  neutrons. a  proton  is a subatomic particle with a positive electric charge. allprotons  are identical.

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