Question

Complete the following radioactive decay problem.
210/84 Po --> 206/82 Pb + ____/____ __

Answer 1

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$  

Your equation is:  

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + ?

It becomes easier to balance the equation if we replace the "?" with an element symbol _x^yZ.  

Then the equation becomes  

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +_x^yZ

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.  

Sum of superscripts: 210 = 206 + y, so y =4.  

Sum of subscripts: 84 = 82 + x, so x = 2.

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

We should recall that $_2^4He$ is an α particle.

Thus, the equation represents the α decay of polonium-210 to lead-206.

The nuclear equation is  

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

Answer 2

Answer:

4/2 Be

210-206=4 and 84-82=2