Question

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What is the nuclide symbol of X? Superscript 231 subscript 94 upper P u. Superscript 235 subscript 90 upper T h. Superscript 239 subscript 94 upper P u. Superscript 231 subscript 90 upper T h. the nuclear equation below.

Answer 1

Answer: The nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

$_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}$

For the given equation :

$^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}$

As the atomic number and mass number must be equal on both sides of the nuclear equation:

$^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}$

Thus the nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Answer 2

Answer:

D. Superscript 231 subscript 90 upper T h.

Explanation: Just for a short answer