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, t_1/2 =10 years

K = 0.693 / t_{1/2 =}  0.693  / 10

K= 0.0693 years¹ Ans

Average life = 1.44 × t_1/2=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 t_1/2 =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×t_1/2

= 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 –

t_1/2=24 days

T=n×t_1/2

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 t_1/2 = 60 days , T  = 180 days

Let initial  reactivity ‘No’ = 100 gm

T=n×t_1/2

n = T / t_1/2

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×t_1/2

90 = 3 × t_1/2

t_1/2 =90/3 = 30 days 

K =0.693 /t_1/2 = 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×t_1/2

48  = 3 × t_1/2

t_1/2 =48/3 = 16  days 

K =0.693 /t_1/2 = 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  , t_1/2 = 13.3 years

T=n×t_1/2

n  = T /  t_1/2  = 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.