radioactive disintegration series

**source : schools.aglasem.com
**

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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.

**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:

_{90}Th^{232} _{88}Ra^{228} _{89}Ac^{228} _{90}Th^{228} _{88}Ra^{224} _{86}Rn^{220} _{84}Po^{216} _{82}Pb^{212} _{83}Bi^{212} _{81}Tl^{208} _{82}Pb^{208}

**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-

_{93}NP^{237} _{91}Pa^{233} _{92}U^{233} _{90}Th^{229} _{88}Ra^{225} _{89}Ac^{225} _{87}Fr^{221} _{85}At^{217} _{83}Bi^{213} _{84}Po^{213} _{82}Pb^{209} _{83}Bi^{209}

**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 U^{238} and ends at Pb^{206}.

**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 _{92}U^{235}. The name actiuranium (ACU) was given to the isotope _{92}U^{235} 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 | _{90}Th^{232} |
_{82}Pb^{208} |
6a and 4 β |

2. Neptunium Series or (4n+1) series | _{93}Np^{237} |
_{83}Bi^{209} |
7a and 4 β |

3. Uranium Series or (4n+2) series | _{92}U^{238} |
_{82}Pb^{206} |
8a and 6 β |

4. Actinium Series or (4n+3) series | _{92}U^{235} |
_{82}Pb^{207} |
7a and 4 β |

**Numericals-**

- To Which disintegration series Rn
^{220}belongs to?

Ans. Mass No. = 220

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

- To Which disintegration Series Pb
^{214}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
_{83}Bi^{209}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
_{92}U^{238}, 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 = _{84}X^{214} 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.

_{90}Th^{234}disintegrates to give_{82}Pb^{206}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 eq^{n} (1)

2x–y=8

2×7–y=8

14 –y = 8

y=6

Ans. No. of α-particles =7

No. of β-particles =6

_{92}U^{235}is the parent of a radioactive decay series which terminates at_{82}Pb^{207}. 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

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