Class 12 Physics Nuclei | Beta decay |

**Beta decay**

- In case of beta decay either electron or positron is emitted.
- Mass number remains the same.
- β
^{−}decay --> Electron is emitted.- Atomic number increases by 1.
- For example:-
^{32}_{15}P à_{16}^{32}S + e^{-}+ ̅ν where ̅ν =anti-neutrino - Q-value for β
^{−}decay:-^{A}_{Z}Xà^{A}_{ (Z+1)}Y + e^{-}+ ̅ν- Initial rest mass energy U
_{i}= [m (^{A}_{Z}X) - Zm_{e}]c^{2}- Where m (
^{A}_{Z}X) = rest mass and Zme = mass of electrons

- Where m (
- Final rest mass energy U
_{f}= [m(^{A}_{ (Z+1)}Y) – (Z+1) m_{e}+ m_{e}] c^{2} - =[m(
^{A}_{ (Z+1)}Y) -Zm_{e}] c^{2} - Therefore Q =U
_{i}– U_{f} - =[m(
^{A}_{Z}X) -Zm_{e}– m(^{A}_{ (Z+1)}Y) + Zm_{e}] c^{2} **Q = [m(**^{A}_{Z}X) - 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:-
^{23}_{11}Na -->_{10}^{22}Ne + e^{+}+ν where ν = neutrino - Q-value for beta (+ive) decay:-
^{A}_{Z}X -->^{A}_{ (Z-1)}Y + e^{+}+ ν- Initial rest mass energy U
_{i}= [m (^{A}_{Z}X) - Zm_{e}]c^{2}- Where m (
^{A}_{Z}X) = rest mass and Zme = mass of electrons

- Where m (
- Final rest mass energy U
_{f}= [m(^{A}_{ (Z-1)}Y) – (Z-1) m_{e}+ m_{e}] c^{2} - Therefore Q =U
_{i}- U_{f} - = [m (
^{A}_{Z}X) - Zm_{e}- m(^{A}_{ (Z-1)}Y) – (Z-1)m_{e}- m_{e}] c^{2} **Q = [m(**^{A}_{Z}X) - m(^{A}_{ (Z-1)}Y) -2 m_{e}] c^{2}

- Example to show whether nuclear reaction can take place or not:-
^{1}_{1}p(proton) à_{0}^{1}n(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
_{0}^{1}n (neutron) à^{1}_{1}p (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 (

** Answer:- **In β

β^{-} Emission of the nucleus (^{23}_{10}Ne)

(^{23}_{10}Ne) --> (^{23}_{11}Na) + e^{-} + ̅ν + Q

It is given that:

Atomic mass m (^{23}_{10}Ne) of = 22.994466 u

Atomic mass m (^{23}_{11}Na) of = 22.989770 u

Mass of an electron, m_{e} = 0.000548 u

Q-value of the given reaction is given as:

Q= [m (^{23}_{10}Ne) + [m (^{23}_{11}Na) +m_{e}]] c^{2}

There are 10 electrons in and 11 electrons in (^{23}_{11}Na). Hence, the mass of the electron is cancelled in the Q-value equation.

Therefore Q = [22.994466 - 22.989770] c^{2}

= (0.004696 c^{2}) u.

But 1u = 9.31 (MeV/c^{2})

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

e^{+} + ^{A}_{Z}X --> ^{A}_{(Z-1)}Y + ν

Show that β^{+}if emission is energetically allowed, electron capture is necessarilyallowed but not vice−versa.

__Answer:-__

Let the amount of energy released during the electron capture process be Q_{1}. Thenuclear reaction can be written as:

e^{+} + ^{A}_{Z}X --> ^{A}_{(Z-1)}Y + ν +Q_{1} (equation(1))

Let the amount of energy released during the positron capture process be Q_{2}. Thenuclear reaction can be written as:

^{A}_{Z}X --> ^{A}_{ (Z-1)} Y + e^{+} + ν +Q_{2} (equation (2))

m_{N} (^{A}_{Z}X) =Nuclear mass of ( ^{A}_{Z}X)

m_{N}(^{A}_{ (Z-1)} Y) = Nuclear mass of (^{A}_{ (Z-1)} Y)

m(^{A}_{Z}X) = Atomic mass of ( ^{A}_{Z}X)

m(^{A}_{ (Z-1)} Y) = Atomic mass of (^{A}_{ (Z-1)} Y)

m_{e} = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

Q_{1}= [m_{N}(^{A}_{Z}X) + m_{e} -m_{N}(^{A}_{ (Z-1)} Y)] c^{2}

= [m (^{A}_{Z}X) -Zm_{e}+ m_{e} –m (^{A}_{ (Z-1)} Y) + (Z-1) m_{e}] c^{2}

= [m (^{A}_{Z}X) -m^{A}_{ (Z-1)} Y] c^{2} (equation (3))

Q-value of the positron capture reaction is given as:

Q_{2}=m_{N}(^{A}_{Z}X) -m_{N}(^{A}_{ (Z-1)} Y) - m_{e}] c^{2}

= [m (^{A}_{Z}X) -Zm_{e}– m (^{A}_{ (Z-1)} Y) + (Z-1) m_{e} - m_{e}] c^{2}

= [m (^{A}_{Z}X) -m^{A}_{ (Z-1)} Y – 2m_{e}] c^{2} (equation (4))

It can be inferred that if Q_{2}> 0, then Q_{1}> 0; Also, if Q_{1}> 0, it does not necessarilymean that Q_{2}> 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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