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

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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
User Mario Plantosar
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