Class 12 Physics
Nuclei
Alpha decay
- In alpha decay α particles are emitted.Daughter nucleus is formed from the parent nucleus.
- The atomic number decreases by 2 and Mass number increases by 4.
- It is a very spontaneous process and happens on its own.
- It occurs only in radionuclides.
- For example:-
- 23892U (unstable) --> 90234Th + 24He
- (Uranium (U) is known as parent nucleus and Thorium (Th) is known as daughter nucleus).
- 24He is α particle.
- General form of alpha decay: -AZX (parent) à(A-4) (Z-2) Y + 24He
- Where Y is daughter nucleus.
- 23892U (unstable) --> 90234Th + 24He
Q-value of alpha decay
- Q-value is a parameter or a characteristic of a nuclear reaction which describeswhether the reaction can take place or not.
- Q-value is defined as difference in the initial mass energy and final mass energy of decayed products.
- Consider the general equation:-
- AZX (parent) -->(A-4) (Z-2) Y + 24
- 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-4) (Z-2) Y – (Z-2) me + m(24He) – 2me ] c2
- Q =Ui - Uf
- =[m (AZX) - Zme - m((A-4) (Z-2) Y + (Z-2) me - m(24He) + 2me]c2
- Q = [m (AZX) - m((A-4) (Z-2) Y- m(24He)]c2
- If Q =(+ive) then equation is energetically allowed.
- => Q>0, (Ui-Uf) >0=> Ui>Uf
- This implies energy of the reactants is more than the energy of the products.
- There is some free energy that is in the form of kinetic energy.
- If Q =(-ive) then equation is not energetically allowed.
Problem: - Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) (22688Ra) and (b) (22086Rn). Given m (22688Ra) =226.0250 u, m (22086Rn) = 220.01337u, m (22286Rn) =222.01750u, m (21684Po) =216.00189u.
Answer:-
(a) Alpha particle decay of (22688Ra) emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.
(22688Ra) --> (22288Ra) + (42He)
Q-value of emitted α-particle = (Sum of initial mass − Sum of final mass)
c2 Where, c = Speed of light It is given that:
m (22688Ra) =226.0250 u, m (22286Rn) =222.01750u, m (42He) =4.002603u
Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2
But 1 u = 931.5 (MeV/c2)
Therefore Q = (0.005297 × 931.5) ≈ 4.94 MeV
Kinetic energy of the α-particle =
(Mass number after decay)/ (Mass number before decay) x Q
= (222)/ (226) x 4.94 = 4.85MeV.
(b) Alpha particle decay of (22086Rn)
(22086Rn) --> (21684Po) + (42He)
It is given that:
Mass of (22086Rn) = 220.01137 u
Mass of (21684Po) = 216.00189 u
Therefore, Q-value = [220.01137 – (216.00189+4.002603)] x931.5
≈ 641 MeV
Kinetic energy of the α-particle = (220-4)/ (220) x 6.41
= 6.29 MeV
Problem: -We are given the following atomic masses:23892U = 238.05079 u
42He = 4.00260 u, 90234Th = 234.04363 u,11H= 1.00783 u, 23791Pa = 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
(a) Calculate the energy released during the alpha decay of 23892U .
(b) Show that 23892U cannot spontaneously emit a proton.
Answer:-
(a) The alpha decay of 23892U is given by equation
AZ X --> (A-4) (Z-2Y + 42He
The energy releasedin this process is given by:
Q = (MU – MTh – MHe) c2
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260) u × c2
= (0.00456 u) c2
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV.
(b) If 23892U spontaneously emits a proton, the decay process would be
23892U --> 23791Pa + 11H
The Q for this process to happen is
= (MU – MPa – MH) c2
= (238.05079 – 237.05121 – 1.00783) u × c2
= (– 0.00825 u) c2
= – (0.00825 u) (931.5 MeV/u)
= – 7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceedspontaneously. We will have to supply energy of 7.68 MeV to a23892U nucleus to make it emit a proton.
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