what is the value of z in the following nuclear reaction? 237 93np→233 91pa+azx

let me rewrite the  equation

²³⁷ ₉₃Np→²³³ ₉₁Pa+^a_zX

 from the above equation neptunium losses 4 from mass number and 2 from atomic number,  this is an alpha decay , therefore the decay equation will be

²³⁷ ₉₃Np→²³³ ₉₁Pa+⁴₂He

 therefore a=4 , z=2 and X=alpha particle.