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 years ^{1} Ans*

*Average life = 1.44 × t _{1/2}=1.44×10*

*Average life = 14.4 years Ans.*

### Question 2) Half life period of U^{234} is 2.5×10^{5} years. In how many years it will remain 25% of original amount?

### Ans.

* Given t _{1/2} =2.5×10^{5} years*

*No = 100 gm*

*N = 25 grams*

*N= N _{0}(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}*

*= 2×2.5×10 ^{5}*

*T=5×10*^{5} years Ans

^{5}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= N _{0}(1/2)^{n}*

*N= 10 (1/2) ^{2}*

*N =2.5 grams*

### Question 4) Half life of_{83} I^{125} 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= N _{0}(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= N _{0}(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

_{1/2 }=90/3 = 30 days

*K =0.693 /t _{1/2} = 0.693 / 30*

*K = 0.0231 days*^{-1}

^{-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= N _{0}(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*

*N _{0} = 100 gm*

*Disintegrated amount = 87.5 gm*

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

*N = 12.5 grams*

*N= N _{0}(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

_{1/2 }=48/3 = 16 days

*K =0.693 /t _{1/2} = 0.693 / 16*

*K = 0.043 days*^{-1} Ans.

^{-1}Ans.

### Question 8) Half life of _{1}H^{3} 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 , N*_{0} = ?

_{0}= ?

*N= N _{0}(1/2)^{n}*

*4 = N _{0} (1/2)^{2}*

*4 = N _{0 }/4*

*N _{0} = 16 kg Ans.*

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

### Solution )

*Initial amount = N _{0} *

*Amount left after 5 half lives ‘N’ = ? *

n = 5

*N= N _{0}(1/2)^{n}*

*N = N _{0} (1/2)^{5}*

* = N _{0 } (1/32) = N_{0} / 32 *

#### N = *N*_{0} / 32

_{0}/ 32