Radioactive Disintegration Series

radioactive disintegration series

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            “A series of regular disintegration starting from an unstable nucleus and ending at a stable nucleus, is known as radioactive disintegration series”.

Or

“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.

4. 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

Numericals-

1. To Which disintegration series Rn²²⁰ belongs to?

Ans. Mass No. = 220

220 is divisible by 4. Hence it is the member of 4n Series.

1. 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.

1. 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.

1. 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

= 92–12+4

= 84

Atomic weight of the new element    = 238–6×4=238-24

= 214

New Element = ₈₄X²¹⁴  Ans.

1. 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.

1. 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.

90–2x+1y= 82

2x–y= 90–82

2x–y=8                        (1)

234–4x = 206  (2)

4x        = 234-206

= 28

x          = 28/4=7

x          = 7

Putting the value of x in eqⁿ    (1)

2x–y=8

2×7–y=8

14 –y = 8

y=6

Ans.     No. of α-particles =7

No. of β-particles =6

207. 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

1. of β–particles emitted =y

92–2x+y×1=82

92–82=2x–y

2x–y =10                     (1)

235–4x=207                (2)

235–207=4x

4x=28

x=28/4=7

x=7

Putting the value of x in eq (1)

2x–y=10

2×7–y=10

14-10=y

y=4

Ans.     No. of α-particles =7

No. of β-particles =4