92238 U has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes:

Question

92238 U has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes: (A)  92234 U (B)  90234 Th (C)  92235 U (D)  93237 Np

Tardigrade Admin · Accepted Answer

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 ⟶_{Z}^{A} X +_{2}^{4} He

From conservation of atomic mass

238 = A + 4

\Rightarrow A = 234

From conservation of atomic number

92 = Z + 2

\Rightarrow Z = 904

So, the resultant nucleus is

_{90}^{234} X

,
i.e.,

_{90}^{234} T h

.
Note: A nuclide below the stability line in uranium decay series disintegrates in such a way that its proton number decreases and its neutron to proton ratio increases. In heavy nuclides this can occur by alpha emission.

Q. $_{92}^{238} U$ has $92$ protons and $238$ nucleons. It decays by emitting an alpha particle and becomes:

$_{92}^{234} U$

$_{90}^{234} T h$

$_{92}^{235} U$

$_{93}^{237} Np$

Q. 92238​U has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes:

92234​U

90234​Th

92235​U

93237​Np

Q. $_{92}^{238} U$ has $92$ protons and $238$ nucleons. It decays by emitting an alpha particle and becomes:

$_{92}^{234} U$

$_{90}^{234} T h$

$_{92}^{235} U$

$_{93}^{237} Np$