Question

Identify the nuclide produced when neptunium-237 decays by alpha emission:

237 93Np→42He + ?

Express your answer as an isotope using prescripts.

Answer 1

`_(91)^(233)\textrm{Pa}`

Step-by-step explanation:

In a nuclear reaction, the total mass and total atomic number remains the same. It is mentioned that the a alpha particle,
`^4_2\textrm{He}` is being emitted.

For the given fission reaction:

`^(237)_(93)\textrm{Np}\rightarrow ^A_Z\textrm{X}+^4_2\textrm{He}`

To calculate A:

Total mass on reactant side = total mass on product side

237 = A + 4

A = 233

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

93 = Z + 2

Z = 91

Hence, the isotopic symbol of unknown element is
`_(91)^(233)\textrm{Pa}`