WBCHSE Class 12 Physics Notes
Atomic Nucleus Displacement Laws Of Radioactive Decay
Soddy and Fajans formulated two laws based on the observations made on radioactive decay(α -decay and β -decay). These are known as Soddy-Fajans’ displacement laws.
Law α -decay
Due to α -the decay of a radioactive nucleus, the mass and charge of the daughter nucleus decrease by 4 and 2 respectively from the parent nucleus
⇒ \(\underset{\mathrm{parent}}{Z^{X^A}} \longrightarrow \underset{\mathrm{daughter}}{Z-2 Y^{A-4}}+{ }_2 \mathrm{He}^4\)
Short Notes on Displacement Laws in Radioactivity
Example: \(\underset{\mathrm{radium}}{{ }_{88} \mathrm{Ra}^{226}}\longrightarrow \underset{\mathrm{radon}}{{ }_{86} \mathrm{Rn}^{222}}+{ }_2 \mathrm{He}^4\)
Law of β -decay:
Due to β – decay of a radioactive nucleus, the mass of the daughter nucleus remains the same
As that of the parent nucleus, the charge of the daughter nucleus is increased by 1. It Is interesting to note that the atomic number of the daughter element increases by 1 because during β -decay a neutron is converted into a proton.
⇒ \(\underset{\text { (parent) }}{Z^{X^A}} \longrightarrow \underset{\text { (daughter) }}{Z+1} Y^A+{ }_{-1} e^0\)
Example: \(\underset{\text { (thrium) }}{{ }_{90} \mathrm{Th}^{234}} \longrightarrow \underset{\text { (protactinium) }}{{ }_{91} \mathrm{~Pa}^{234}}+{ }_{-1} e^0\)
Note that the daughter nucleus is an isobar of the parent nucleus
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Rule of emission of γ-ray:
Schematically represents the emission of γ -ray. Due to the emission of a or β -participation cle from the parent nucleus at ground state energy level X, the daughter nucleus stays in an excited energy level Y. To be stable, γ -radiation takes place taking the daughter nucleus to its ground state energy level Y. Thus γ -radiation involves only a transition in energy level of the daughter nucleus and the structure of the nucleus does not change. Hence there is no change in atomic mass number or atomic number
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WBBSE Class 12 Radioactive Decay Notes
Conservation laws of mass number and atomic number
In addition to the well-established laws of conservation of momentum, angular momentum, and mass energy, two more conservation laws are to be specially mentioned in the case of radioactive decay.
- Conservation of mass number: Radioactive decay does not bring about any change in the total number of neutrons and protons. So, the mass number remains unchanged.
- Conservation of atomic number (proton number): There is no change in the total number of protons in the reactant (parent) and the product (daughter nucleus + emitted particle). So atomic number remains the same. These conservation laws also hold good in the case of the artificial transmutation of elements.
Disintegration energy
In both α and β – decay, the mass of the products (the daughter nucleus and emitted particle) is found to be less than the mass of the parent nucleus. As per Einstein’s mass-energy equivalence, this lost mass is transferred to energy and this energy is called disintegration energy. Most of this energy is carried away by the α or β -particles, as they are lighter than the daughter nucleus. Thus nuclear disintegration produces high energy α -particle or γ -particle.
Displacement laws of radioactive decay
Atomic Nucleus Displacement Laws Of Radioactive Decay Numerical Examples
Example 1: How many α and β -particles are emitted when U-238 changes to Pb-206 due to radioactivity? Atomic numbers of U-238 and Pb-206 are 92 and 82 respectively.
Solution:
As per the displacement rule, the loss of mass number due to the emission of an α -particle is 4 and that due to the emission of β – particle is nil. Also, the decrease in atomic number due to α – emission is 2, and the increase in atomic number due to β – emission is +1.
Let x and y be the required numbers of α and β -emissions for the transmutation.
∴ 4x+0 = 238-206 = 32
x = \(\frac{32}{4}\) = 8
Again, the reduction in atomic number due to a -emission = 2x and the increase in atomic number due to /9 -emission =y.
Total reduction in atomic number = 2x- y
Now, according to the question,
2x – y = 92 – 82 = 10
y = 16 – 10 = 6
Hence, 8 α -particles and 6 β -particles are emitted
Conceptual Questions on Types of Radioactive Decay
Example: 2. decays by emitting successively 8 α -particles and 6 β -particles. Determine the mass number and atomic number of the new element and express it in a symbol.
Solution:
Using the Soddy-Fajans’ displacement rule, Loss in mass number due to α -emission = 8 × 4 = 32, and there is no change in mass number due to β -emission. Hence, the mass number of the element formed
Due to the emission of 8 a -particles the decrease in atomic number
= 8 × 2 = 16
Now, due to the emission of 6 -particles the increase in the atomic number
6 ×1 = 6
The atomic number of a new element
= 92- (16 – 6) = 92 – 10
= 82
The atomic number is 82, the element formed is lead (Pb) and the symbolic representation is 92Pb206
Radioactive decay class 12 notes Practice Problems on Half-Life and Decay Constants
Example 3. 86Pb222 → 84Pb210 Determine how many α -particles end β -particles have been emitted in the above reaction:
Solution:
The reduction in mass number = 222- 210 = 12
This reduction in mass number can only be caused by to emission of a -particles. Now since the mass of an a -particle is 4, the number of α -particles emitted = \(\frac{12}{4}\)= 3
Again, due to the emission of 3 a -particles the decrease in atomic number =2 × 3 = 6
In the reaction the decrease in atomic number = 86 – 84 = 2
∴ Due to the emission of β -particles increase in atomic number 6 – 2 = 4
Since due to the emission of β -particle atomic number increases 4 by 1, the number of β – particles emitted \(\frac{4}{1}\) = 4