Radon 222 86 Rn is the start of a decay chain that forms bismuth 214 83 Bi by alpha and beta emission.

Question 2

Radon ²²²₈₆Rn is the start of a decay chain that forms bismuth ²¹⁴₈₃Bi by alpha and beta emission.

For the decay of each nucleus of radon, how many α−particles and β−particles are emitted?

α−particles     β−particles

A                      1                      1

B                      2                      1

C                     1                      2

D                     2                      2

Reference: Past Exam Paper – June 2012 Paper 12 Q40

Solution:

Answer: B.

α-particle = ⁴₂He and β-particle = ⁰_-1e

Let ‘m’ be the number of α-particles emitted and ‘n’ be the number of β-particles emitted.

²²²₈₆Rn        -->       ²¹⁴₈₃Bi +          m ⁴₂He            +          n  ⁰_-1e

The mass number and proton number should be conserved.

Mass number: 222 = 214 + 4m + 0n

4m = 8 giving m = 2    (so, 2 α-particles emitted)

Proton number: 86 = 83 + 2m – 1n

2m – n = 3

n = 4 – 3 = 1                since m = 2 (from above)

So, 1 β-particle is emitted.