Class 12 Physics Nuclei Alpha decay

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.

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.

(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.

(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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