|Class 12 Physics Nuclei||Beta decay|
Beta-decay: neutrino and anti-neutrino
Problem:- The nucleus (2310Ne) decays by β- emission. Write down the β decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:m (2310Ne) = 22.994466 u,m (2311Na) = 22.989770 u.
Answer:- In β- emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
β- Emission of the nucleus (2310Ne)
(2310Ne) --> (2311Na) + e- + ̅ν + Q
It is given that:
Atomic mass m (2310Ne) of = 22.994466 u
Atomic mass m (2311Na) of = 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
Q= [m (2310Ne) + [m (2311Na) +me]] c2
There are 10 electrons in and 11 electrons in (2311Na). Hence, the mass of the electron is cancelled in the Q-value equation.
Therefore Q = [22.994466 - 22.989770] c2
= (0.004696 c2) u.
But 1u = 9.31 (MeV/c2)
Therefore Q= (0.004696 x 931.5) =4.374MeV.
The daughter nucleus is too heavy as compared to e- and ̅ν. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.
Problem:- For the β+(positron) emission from a nucleus, there is another competing process knownas electron capture (electron from an inner orbit, say, the K−shell, is captured by thenucleus and a neutrino is emitted).
e+ + AZX --> A(Z-1)Y + ν
Show that β+if emission is energetically allowed, electron capture is necessarilyallowed but not vice−versa.
Let the amount of energy released during the electron capture process be Q1. Thenuclear reaction can be written as:
e+ + AZX --> A(Z-1)Y + ν +Q1 (equation(1))
Let the amount of energy released during the positron capture process be Q2. Thenuclear reaction can be written as:
AZX --> A (Z-1) Y + e+ + ν +Q2 (equation (2))
mN (AZX) =Nuclear mass of ( AZX)
mN(A (Z-1) Y) = Nuclear mass of (A (Z-1) Y)
m(AZX) = Atomic mass of ( AZX)
m(A (Z-1) Y) = Atomic mass of (A (Z-1) Y)
me = Mass of an electron
c = Speed of light
Q-value of the electron capture reaction is given as:
Q1= [mN(AZX) + me -mN(A (Z-1) Y)] c2
= [m (AZX) -Zme+ me –m (A (Z-1) Y) + (Z-1) me] c2
= [m (AZX) -mA (Z-1) Y] c2 (equation (3))
Q-value of the positron capture reaction is given as:
Q2=mN(AZX) -mN(A (Z-1) Y) - me] c2
= [m (AZX) -Zme– m (A (Z-1) Y) + (Z-1) me - me] c2
= [m (AZX) -mA (Z-1) Y – 2me] c2 (equation (4))
It can be inferred that if Q2> 0, then Q1> 0; Also, if Q1> 0, it does not necessarilymean that Q2> 0.
In other words, this means that if β+emission is energetically allowed, then the electroncapture process is necessarily allowed, but not vice-versa. This is because the Q-valuemust be positive for an energetically-allowed nuclear reaction.