NCERT Class XII Chapter
Nuclei
Questions With Solutions
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Question : 13
Total: 31
The radionuclide
C
C
B + e + T 1 ∕ 2
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m(
C ) (
B )
Calculate Q and compare it with the maximum energy of the positron emitted
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m
Calculate Q and compare it with the maximum energy of the positron emitted
Solution:
The given equation,
C
B + e +
The Q value is, Q =[ m (
C ) − 6 m e − m (
B ) + 5 m e − m e ] c 2
Q = [11.011434 – 11.009305 – 2 × 0.000548]c 2
931.5 MeV ≈ 0.96 MeV
As we know that different positrons comes out with different possible energies shared between daughter nucleus and positron.
So, the Q value of reaction is almost same as the maximum energy of positron emitted.
The Q value is, Q =
Q = [11.011434 – 11.009305 – 2 × 0.000548]
931.5 MeV ≈ 0.96 MeV
As we know that different positrons comes out with different possible energies shared between daughter nucleus and positron.
So, the Q value of reaction is almost same as the maximum energy of positron emitted.
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