Complete the following radioactive decay problem. 238\92u arrow 4/2he 234/90th arrow 234/91pa ( ? ) 4/2he 1/1p 1/0n 0/1e 0/-1e

Here we have to complete the given radioactive decay reaction:

²³⁸₉₂U → ⁴₂He + ²³⁴₉₀Th → ²³⁴₉₁Pa ? ⁴₂He + ₁¹P + ¹₀n + ⁰₁e + ₋₁⁰e.

The completed radioactive reaction is

²³⁸₉₂U → ⁴₂He + ²³⁴₉₀Th → ²³⁴₉₁Pa + ⁴₂He + ₋₁⁰e (electron).

In any radioactive reaction the mass number and the atomic number of the reactant and product will be same respectively.

The mass number and atomic number of the reactant uranium (U) is 238 and 92 respectively. Thus in the last reaction if ⁴₂He is eliminated then the mass number of the product 234 + 4 - 0 = 238.

The atomic number will be 91 + 2 - 1 = 92.

Thus the missing particle is electron (₋₁⁰e)

i think   the answer is c hope i'm right

the correct answer is:

a. formation of precipitation