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

Options

A. `._(92)^(234)U`
B. `._(90)^(234)Th`
C. `._(92)^(235)U`
D. `._(93)^(237)Np`

Answer

b

Explanation

As a general rule in any decay sum of mass number `A` and atomic number `Z` must be same on both sides.
Let the daughter nucleus be `._(Z)^(A)X`
So, reaction can be shown as
`._(92)^(238)U rarr._(Z)^(A)X+He`
From conservation of atomic mass
`238= A+4 implies A= 234`
From conservation of atomic number
`92= Z+2`
`implies Z= 90`
So, the resultant nucleus is `._(90)^(234)X` i.e., `._(90)^(234)Th`.