Determine the value of x from the nuclear reaction below, $^{222}_{86}Rn$ → $^{x}_{88}Ra$ + 2β + energy

Analyze the question

We have a nuclear reaction: Rn-222 (atomic number 86) → Ra-x (atomic number 88) + 2β + energy. We need the mass number x.

Concept

In any nuclear reaction, two quantities are conserved:

• Mass number (total nucleons), the top number
• Atomic number (total charge), the bottom number

A beta particle (β⁻) is an electron: mass number 0, charge −1, written as ⁰₋₁e.

Solution

Charge check: the atomic number goes from 86 to 88, an increase of 2. Two β⁻ particles each carry charge −1, so the right side has 88 + 2(−1) = 86. This matches the left side. So the equation is consistent with two β⁻ particles.

Mass number: beta particles have negligible mass (mass number 0), so:

222 = x + 2(0) x = 222

So the reaction is:

²²²₈₆Rn → ²²²₈₈Ra + 2 ⁰₋₁e + energy

Answer

x = 222, which is option B.

Note

In real life, Ra-222 isn't produced this way from Rn-222, but the question is a conservation exercise, so just apply the rules.

Exam tip

Beta emission changes the atomic number (+1 for each β⁻) but leaves the mass number unchanged. Alpha emission lowers the mass number by 4 and the atomic number by 2. Always check that the top numbers balance and the bottom numbers balance.