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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