Class 12 Physics Nuclei Beta decay

Beta decay

• In case of beta decay either electron or positron is emitted.
• Mass number remains the same.
• βdecay --> Electron is emitted.
• Atomic number increases by 1.
• For example:- 3215P à1632S + e- + ̅ν  where  ̅ν =anti-neutrino
• Q-value for βdecay:-
• AZA (Z+1) Y + e-+ ̅ν
• Initial rest mass energy Ui = [m (AZX) - Zme]c2
• Where m (AZX) = rest mass and Zme = mass of electrons
• Final rest mass energy Uf = [m(A (Z+1) Y) – (Z+1) me + me ] c2
• =[m(A (Z+1) Y) -Zme] c2
• Therefore Q =Ui– Uf
• =[m(AZX) -Zme – m(A (Z+1) Y) + Zme] c2
• Q = [m(AZX) - m(A (Z+1) Y)]

Example:-

• β+decay à Positron(same as electron but with (+ ive) charge) is emitted.
• Atomic number decreases by 1.
• For example:- 2311Na -->1022Ne + e++ν where ν = neutrino
• Q-value for beta (+ive) decay:-
• AZX -->A (Z-1) Y + e+ + ν
• Initial rest mass energy Ui = [m (AZX) - Zme]c2
• Where m (AZX) = rest mass and Zme = mass of electrons
• Final rest mass energy Uf = [m(A (Z-1) Y) – (Z-1) me + me ] c2
• Therefore Q =Ui - Uf
• = [m (AZX) - Zme - m(A (Z-1) Y) – (Z-1)me - me] c2
• Q = [m(AZX) - m(A (Z-1) Y) -2 me] c2
• Example to show whether nuclear reaction can take place or not:-
• 11p(proton) à01n(neutron) + e+ + ν. The mass of neutron is greater than the mass of proton.
• The Q-value = (-ive) as a result the above reaction is not possible.This means conversion of stable proton to neutron is not allowed.
• Consider 01n (neutron) à11p (proton) + e- + ̅ν
• The Q-value =(+ive) as a result above reaction is possible.This means conversion of neutron to proton is allowed.

Beta-decay: neutrino and anti-neutrino

• Neutrino (ν) and anti-neutrino ( ̅ν) are neutral particles.
• They also have negligible mass.
• They have extremely high penetration power.
• They have extremely weak interaction with matter.

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.

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