Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) 22688Ra and (b)22086Rn. Given m(22688Ra) = 226.02540 u, m(22286Rn) = 222.01750 u, m(22086Rn) = 220.01137 u, m(21684Po) = 216.00189 u.
Answers
Answer:
The beta decay equation is as shown.
10
23
Ne→
11
23
Na+e
−
+Q
Q=Δm× 931.5 MeV
Here, Δm=m
N
(
10
23
Ne)−m
N
(
11
23
Na)−m
e
= mass defect.
Δm=[m
N
(
10
23
Ne)−m
N
(
11
23
Na)−m
e
]=22.994466−22.089770=0.904696 amu
Here, the masses used are masses of nuclei and not of atoms. Note: m
e
has been cancelled.
Q=Δm× 931.5 MeV =0.904696×931.5=4.37MeV.
The maximum kinetic energy of the electron (max E
e
) = Q = 4.37 MeV.
Explanation:
The beta decay equation is as shown.
10
23
Ne→
11
23
Na+e
−
+Q
Q=Δm× 931.5 MeV
Here, Δm=m
N
(
10
23
Ne)−m
N
(
11
23
Na)−m
e
= mass defect.
Δm=[m
N
(
10
23
Ne)−m
N
(
11
23
Na)−m
e
]=22.994466−22.089770=0.904696 amu
Here, the masses used are masses of nuclei and not of atoms. Note: m
e
has been cancelled.
Q=Δm× 931.5 MeV =0.904696×931.5=4.37MeV.
The maximum kinetic energy of the electron (max E
e
) = Q = 4.37 MeV.
Answer:
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