According to this decay series chart, what will be the total number of decay particles emitted when one uranium-238 atom decays to form one radon-222 atom?
a) 4 alpha particles and 2 beta particles
b) 1 alpha particle and 1 beta particle
c) 3 alpha particles and 2 positrons
d) 3 alpha particles and 3 gamma rays
Decay series chart: https://curriculum.kcdistancelearning.com/courses/CHEMx-HS-U10/b/assessments/n-nuclear_chemistry_unit_exam/diagram_38_600x785.jpg
what will be the total number of decay particles emitted when one uranium-238 atom decays to form one radon-222 atom?
2016-06-12 2:45 pm
回答 (1)
2016-06-12 2:59 pm
Denote mass number as A, and atomic number as Z.
Let n and m be the number of α particle (helium nuclei) and that of β particles (electrons) emitted when one uranium-238 atom decays to one radon-222 atom.
U(Z=92,A=238) → Rn(Z=86,A=222) + n α-particle(Z=2,A=4) + m β-particle(Z=-1,A=0)
The total mass numbers (A) are unchanged in the decay :
238 = 222 + 4n + 0m
n = 4
The total atomic numbers (Z) are unchanged in the decay :
92 = 86 + 2n + (-1)m
92 = 86 + 2×4 - m
m = 2
...... The answer is : a) 4 α-particles and 2 β-particles
Let n and m be the number of α particle (helium nuclei) and that of β particles (electrons) emitted when one uranium-238 atom decays to one radon-222 atom.
U(Z=92,A=238) → Rn(Z=86,A=222) + n α-particle(Z=2,A=4) + m β-particle(Z=-1,A=0)
The total mass numbers (A) are unchanged in the decay :
238 = 222 + 4n + 0m
n = 4
The total atomic numbers (Z) are unchanged in the decay :
92 = 86 + 2n + (-1)m
92 = 86 + 2×4 - m
m = 2
...... The answer is : a) 4 α-particles and 2 β-particles
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