Test Index

NCERT Class XII Chapter
Nuclei
Questions With Solutions

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Question : 14
Total: 31
The nucleus
23
‌
10
Ne decays by β– emission. Write down the β– decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m (
23
‌
10
Ne) = 22.994466 amu, m (
23
‌
11
Na) = 22.989770 amu
Solution:  
The β– decay of
23
‌
10
Ne may be explained as
23
‌
10
Ne →
23
‌
11
Na+
0
‌
−1
e+
−
v
+Q
The expression or the kinetic energy released may be written as
Q = [m(
23
‌
10
Ne)−m(
23
‌
11
Na)−me]c2 = [m(
23
‌
10
Ne)−m(
23
‌
11
Na)]c2
≈ [22.994466 – 22.989770] × 931.5 MeV
≈ 0.004696 × 931.5 MeV = 4.374 MeV
As
23
‌
11
Na is massive, the kinetic energy released is mainly shared by electron-positron pair. When the neutrino carries no energy, the electron has a maximum kinetic energy equal to 4.374 MeV.
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