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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Friday, November 3, 2017

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.

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