half life

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### 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

### 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

t1/2=24 days

T=n×t1/2

n =48/24 =2

N= N0(1/2)n

N= 10 (1/2)2

### 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

### 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

### Solution )

No = 16 atoms

N = ?

n = 4

N= N0(1/2)n

N = 16 (1/2)4

=  16 (1/16) = 1

### 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

### 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= N0(1/2)n

4  = N0 (1/2)2

4  = N0 /4

N0 = 16 kg Ans.

### 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