Question

Balanced nuclear equation forces the alpha decay Po-210

Answer 1

`_(84)^(210)\text{Po} \longrightarrow \, _(82)^(206)\text{Pb} \, + \, _(2)^(4)\text{He}`

Step-by-step explanation:

Your unbalanced nuclear equation is:

`_(84)^(210)\text{Po} \longrightarrow \, _(x)^(y)\text{Z} + \, _(2)^(4)\text{He}`

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

Then

84 = x + 2, so x = 84 - 2 = 82

210 = y + 4, so y = 210 - 4 = 206

Element 82 is lead, so the nuclear equation becomes

`_(84)^(210)\text{Po} \longrightarrow \, _(82)^(206)\text{Pb} \, + \, _(2)^(4)\text{He}`