radioactive disintegration series
source : schools.aglasem.com
“A series of regular disintegration starting from an unstable nucleus and ending at a stable nucleus, is known as radioactive disintegration series”.
“All the disintegration steps involved in the formation of a non-radioactive element from a radioactive element constitute the disintegration series.”
Types of Disintegration Series- There are four disintegration series.
- i) Thorium Series or 4n series- The mass number of all the elements of this series are divisible by 4. The various steps of this series are given below:
90Th232 88Ra228 89Ac228 90Th228 88Ra224 86Rn220 84Po216 82Pb212 83Bi212 81Tl208 82Pb208
2) Neptunium Series or (4n+1) Series- The mass number of the elements of this series are divided by 4, the remainder is always one. The elements formed in this series are not found in nature. It is called artificial series. The various steps of this series are as-
93NP237 91Pa233 92U233 90Th229 88Ra225 89Ac225 87Fr221 85At217 83Bi213 84Po213 82Pb209 83Bi209
3) Uranium Series or (4n+2) Series- The mass number of the elements of this series are divided by 4, remainder is always 2, This series starts from U238 and ends at Pb206.
4) Actiuranium Series or (4n+3) series- The mass number of the elements of this series are divided by 4, remainder is always 3. This series starts from 92U235. The name actiuranium (ACU) was given to the isotope 92U235 earlier. Therefore the series is called actinium series
|Disintegration Series||Parent Element||Last Stable element||Total No.l of a-and β-particles emitted|
|1. Thorium Series or 4n series||90Th232||82Pb208||6a and 4 β|
|2. Neptunium Series or (4n+1) series||93Np237||83Bi209||7a and 4 β|
|3. Uranium Series or (4n+2) series||92U238||82Pb206||8a and 6 β|
|4. Actinium Series or (4n+3) series||92U235||82Pb207||7a and 4 β|
- To Which disintegration series Rn220 belongs to?
Ans. Mass No. = 220
220 is divisible by 4. Hence it is the member of 4n Series.
- To Which disintegration Series Pb214 belongs to?
Ans. Mass No. = 214
on dividing mass number (214) by 4, remainder is 2. Hence it belongs to (4n+2) series.
- To which disintegration series 83Bi209 belongs to ?
Ans. Mass No. =209
on dividing mass number (209) by 4, remainder is 1. Hence it belongs to (4n+1) series.
- If 6 alpha and 4 beta particles are emitted from 92U238, what would be the atomic number and the mass no. of new element?
Ans. When one α-particle is emitted, the atomic number is decreased by 2 & the Atomic weight is decreased by 4. When one β-particle is emitted, the atomic number is increased by one and there is no change in Atomic weight.
Atomic number of the new element = 92–6×2+4×1
Atomic weight of the new element = 238–6×4=238-24
New Element = 84X214 Ans.
- A radioactive element A decays as follows-
identify the isotopes and isobars among A,B,C&D
Ans. Suppose Atomic No. of A is × & At Wt is Y
Atomic no. of A&D is same therefore A&D are isotope
Atomic weight of B,C&D is same, therefore B,C& D are isobars.
- 90Th234disintegrates to give 82Pb206 as the final product. How many alpha & beta particles are emitted during this process?
Ans. Suppose the number of a-and β-particles emitted is x& y respectively. When one a-particle is emitted, the atomic number is decreased by 2 & Atomic weight is decreased by 4. When one β-particle is emitted, the atomic number is increased by one and there is no change in atomic weight.
234–4x = 206 (2)
4x = 234-206
x = 28/4=7
x = 7
Putting the value of x in eqn (1)
14 –y = 8
Ans. No. of α-particles =7
No. of β-particles =6
- 92U235 is the parent of a radioactive decay series which terminates at 82Pb207. Calculate the total no. of a-and β-particles emitted in the series
Ans. Suppose no. of a–particles emitted =x
- of β–particles emitted =y
2x–y =10 (1)
Putting the value of x in eq (1)
Ans. No. of α-particles =7
No. of β-particles =4