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kondor19780726 [428]
3 years ago
11

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.
Chemistry
1 answer:
Leto [7]3 years ago
5 0

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}

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