`._(92)^(238)U` has 92 protons and `238` nucleons. It decays by emitting an alpha particle and becomes:

Options

A. `._92^234 U`
B. `._92^235 U`
C. `._90^234 Th`
D. `._93^237 Np`

Answer

C

Explanation

( c) Let the daughter nucleus be `._Z^A X`. So, reaction can be shown as
`._92^238 U rarr ._Z^A X + ._2^4 He`
From conservation of atomic mass
`238 = A + 4`
`rArr A = 234`
From conservation of atomic number
`92 = Z + 2`
`rArr Z = 90`
So, the resultant nucleus is `._90^234 X`, i.e., `._90^234 Th`.