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Half Life -Numerical

half life

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Question 1 ) Half life period of a radioactive element is 10 years. Calculate its disintegration constant & average life?

Ans.

Given, t1/2 =10 years

K = 0.693 / t1/2 =  0.693  / 10

K= 0.0693 years1 Ans

Average life = 1.44 × t1/2=1.44×10

Average life = 14.4 years Ans.

Question 2) Half life period of U234 is 2.5×105 years. In how many years it will remain 25% of original amount?

Ans.

Given t1/2 =2.5×105 years

No = 100 gm

N =  25 grams

N= N0(1/2)n

25 = 100 (1/2)n

25 / 100 =(1/2)n

(1/4 ) =(1/2)n

(1/2)2 =(1/2)n

n=2

T=n×t1/2

= 2×2.5×105

T=5×105 years Ans

Question  3)  10 gram thorium remains 5 grams in 24 days . How much of thorium remains in 48 days ?

Ans –

t1/2=24 days

T=n×t1/2

n =48/24 =2

N= N0(1/2)n

N= 10 (1/2)2

N  =2.5 grams

Question 4) Half life of83 I125 is 60 days.How much its radioactivity remains after 180 days ?

Solution )

Given t1/2 = 60 days , T  = 180 days

Let initial  reactivity ‘No’ = 100 gm

T=n×t1/2

n = T / t1/2

n =180 / 60 = 3

N= N0(1/2)n

N  = 100 (1/2)3

    =100 (1/8) = 12.5

N = 12.5 % Ans.

Question 5) The activity of a radioactive element remains 12.5% in 90 days. Calculate the disintegration constant of the element ?

Solution )

Given T  = 90 days

No = 100 gm

N = 12.5 grams

N= N0(1/2)n

12.5 = 100 (1/2)n

12.5 / 100 =(1/2)n

(1/8 ) =(1/2)n

(1/2)3 =(1/2)n

n=3

T=n×t1/2

90 = 3 × t1/2

t1/2 =90/3 = 30 days 

K =0.693 /t1/2 = 0.693 / 30

K = 0.0231 days-1

Question 6) Starting with 16 atoms of a radioactive element , how many atoms are left after 4 half lives?

Solution )

No = 16 atoms

N = ? 

n = 4

N= N0(1/2)n

N = 16 (1/2)4

  =  16 (1/16) = 1 

N = 1 atom Ans.

Question 7) Thorium is 87.5 % disintegrated in 48 days . Calculate the disintegration constant ?

Solution )

Given T  = 48 days

N0 = 100 gm

Disintegrated amount = 87.5 gm

Amount of Thorium left ‘N’ = 100 – 87.5 = 12.5 gm

N = 12.5 grams

N= N0(1/2)n

12.5 = 100 (1/2)n

12.5 / 100 =(1/2)n

(1/8 ) =(1/2)n

(1/2)3 =(1/2)n

n=3

T=n×t1/2

48  = 3 × t1/2

t1/2 =48/3 = 16  days 

K =0.693 /t1/2 = 0.693 / 16

K = 0.043 days-1 Ans.

Question 8) Half life of 1H3 is 13.3 years. How much of 1H3 should be taken initially so that it remains 4 Kg after 26.6 years ?

Solution )

n= ? , T = 26.6 years  , t1/2 = 13.3 years

T=n×t1/2

n  = T /  t1/2  = 26.6 / 13.3 = 2 half lives

N = 4 Kg ,  N0 = ? 

N= N0(1/2)n

4  = N0 (1/2)2

4  = N0 /4

N0 = 16 kg Ans.

Question 9 ) How much of a radioactive element is left after 5 half lives ?

Solution )

Initial amount = N0 

Amount left after 5 half lives ‘N’ = ? 

n = 5

N= N0(1/2)n

N =  N0 (1/2)5

  =  N0  (1/32) = N0 / 32 

N = N0 / 32

Amount left after 5 half lives is 1/32 of initial amount . Ans.

Saroj Bhatia: Dr. Saroj Bhatia is an Ph.D in chemistry who has been teaching chemistry for over a decade. Currently she is a respected principal of a renowned college in her hometown. She took this medium for online users. Her proudest achievement is helping people learn chemistry.
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