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:-
^{238}_{92}U (unstable) -->_{90}^{234}Th +_{2}^{4}He- (Uranium (U) is known as parent nucleus and Thorium (Th) is known as daughter nucleus).
_{2}^{4}He is α particle.

- General form of alpha decay: -
^{A}_{Z}X (parent) à^{(A-4)}_{ (Z-2)}Y +_{2}^{4}He- 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:-
^{A}_{Z}X (parent) -->^{(A-4)}_{ (Z-2)}Y +_{2}^{4}- 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-4)}_{ (Z-2)}Y – (Z-2) m_{e}+ m(_{2}^{4}He) – 2m_{e}] c^{2} - Q =U
_{i}- U_{f} - =[m (
^{A}_{Z}X) - Zm_{e}- m(^{(A-4)}_{ (Z-2)}Y + (Z-2) m_{e}- m(_{2}^{4}He) + 2m_{e}]c^{2} **Q = [m (**^{A}_{Z}X) - m(^{(A-4)}_{ (Z-2)}Y- m(_{2}^{4}He)]c^{2}

**If Q =(+ive) then equation is energetically allowed.**- => Q>0, (U
_{i}-U_{f}) >0=> U_{i}>U_{f} - 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.

- => Q>0, (U
**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) (

__Answer:-__

(a) Alpha particle decay of (^{226}_{88}Ra) 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.

(^{226}_{88}Ra) --> (^{222}_{88}Ra) + (^{4}_{2}He)

Q-value of emitted α-particle = (Sum of initial mass − Sum of final mass)

c^{2} Where, c = Speed of light It is given that:

m (^{226}_{88}Ra) =226.0250 u, m (^{222}_{86}Rn) =222.01750u, m (^{4}_{2}He) =4.002603u

Q-value = [226.02540 − (222.01750 + 4.002603)] u c^{2}

= 0.005297 u c^{2}

But 1 u = 931.5 (MeV/c^{2})

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 (^{220}_{86}Rn)

(^{220}_{86}Rn) --> (^{216}_{84}Po) + (^{4}_{2}He)

It is given that:

Mass of (^{220}_{86}Rn) = 220.01137 u

Mass of (^{216}_{84}Po) = 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:

^{4}_{2}He = 4.00260 u,_{ 90}^{234}Th = 234.04363 u,_{1}^{1}H= 1.00783 u, ^{237}_{91}Pa = 237.05121 u

Here the symbol Pa is for the element protactinium (Z = 91).

(a) Calculate the energy released during the alpha decay of ^{238}_{92}U .

(b) Show that ^{238}_{92}U cannot spontaneously emit a proton.

__Answer:-__

(a) The alpha decay of ^{238}_{92}U is given by equation

^{A}_{Z} X --> ^{(A-4)}_{ (Z-2}Y + ^{4}_{2}He

The energy releasedin this process is given by:

Q = (M_{U} – M_{Th} – M_{He}) c^{2}

Substituting the atomic masses as given in the data, we find

Q = (238.05079 – 234.04363 – 4.00260) u × c^{2}

= (0.00456 u) c^{2}

= (0.00456 u) (931.5 MeV/u)

= 4.25 MeV.

(b) If ^{238}_{92}U spontaneously emits a proton, the decay process would be

^{238}_{92}U --> ^{237}_{91}Pa + ^{1}_{1}H

The Q for this process to happen is

= (M_{U} – M_{Pa} – M_{H}) c^{2}

= (238.05079 – 237.05121 – 1.00783) u × c^{2}

= (– 0.00825 u) c^{2}

= – (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 a^{238}_{92}U nucleus to make it emit a proton.

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