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.

Answer 1

its d

Step-by-step explanation:

edg

Answer 2

`\rm_(90)^(231)\text{Th}`

Step-by-step explanation:

The unbalanced nuclear equation is

`\rm _(92)^(235)\text{U} \longrightarrow \, _(2)^(4)\text{He} + X`

Let's write X as a nuclear symbol.

`\rm _(92)^(235)\text{U} \longrightarrow \, _(2)^(4)\text{He} + _(Z)^(A)\text{X}`

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

92 = 2 + Z , so Z = 92 - 2 = 90

And your nuclear equation becomes

`\rm _(92)^(235)\text{U} \longrightarrow \, _(2)^(4)\text{He} +\, _(90)^(231)\text{X}`

Element 90 is thorium, so

`\rm X = _(90)^(231)\text{Th}`