Unit#20: Nuclear Structures - Solved Problems/Numericals

                           

1. A living plant contains approximately the same isotopic abundance of C-14 as does atmospheric carbon dioxide. The observed rate of decay of C-14 from a living plant is 15.3 disintegrations per minute per gram of carbon. How much disintegration per minute per gram of carbon will be measured from a 12900-year-old sample? (The half-life of C-14 is 5730 years.) 

 DATA:

Initial Activity=A₁=15.3disintegration/min

Time of sample=t=12900 years

 Half life of C-14=T_1/2 =5730 years

Present Activity = A₂=?

SOLUTION:

 First we will calculate the number of half lives during the given time

         No.of Half lives=n= t /T_1/2 =12900/5730 = 2.2513

Now,we will calculate the remaining fractions(x) 

x = (1/2)ⁿ =(1/2)^2.2513

x = 0.21

Now, Present activity is given by

   A₂ = x × A₁ = 0.21×15.3 =3.2 disintegration/min

RESULT: The fractions remain after this time is 0.21 and activity will be 3.2 disintegration/min.

 

2. The smallest C-14 activity that can be measured is about 0.20%. If C-14 is used to date an object, the object must have died within how many years? 

 DATA:

Smallest Activity = A= ?

Percentage of sample = x = 0.2%

Half life of C-14 = T_1/2 = 5730 years

 SOLUTION:

First we will calculate fraction of sample left

         x=0.2/100=0.002

Now,we will calculate the number of half llives

x = (1/2)ⁿ 

0.002 = (1/2)ⁿ 

Taking Log O.B.S

log⁡ 0.002=log⁡ (1/2)ⁿ 

Now using property

    log⁡ aⁿ = n log a

we get

log⁡ 0.002=n log⁡(1/2)

  -  2.6989  = n(-0.3010)

n= (-2.6989)/(-0.3010)=8.966

Now, the time required will be given by

       t = n×T_1/2 =8.966×5730=51375 years

RESULT: The object must have died before 51375 years.

 

3. How long will it take for 25% of the C-14 atoms in a sample of C-14 to decay? 

 DATA:

Time required=t=?

Percentage of sample lost= x =25%

Half life of C-14= T_1/2 =5730 years

SOLUTION:

 First we will calculate fraction of sample left

         x=100-25=75%  or 0.75

Now,we will calculate the number of half llives

x = (1/2)ⁿ 

0.75 = (1/2)ⁿ 

Taking Log O.B.S

log⁡0.75=log⁡(1/2)ⁿ 

Now using property

         log⁡ aⁿ = n log a

we get

log⁡0.75=n log⁡(1/2)

  -  0.1249  =n(-0.3010)

n= (-0.1249)/(-0.3010) = 0.415

Now, the time required will be given by

       t=n×T_1/2 = 0.415×5730 = 2378 years

RESULT: The time required to lost 25% of sample is 2378 years.

4. The carbon-14 decay rate of a sample obtained from a young tree is 0.296 disintegration per second per gram of the sample. Another wood sample prepared from an object recovered at an archaeological excavation gives a decay rate of 0.109 disintegration per second per gram of the sample. What is the age of the object?

 DATA:

Activity of young sample= A₂ =0.296 disintegrations /s.gm

Activity of old sample=A₁=0.109 disintegrations /s.gm

Age of sample = t = ?

Half life of C-14= T_1/2 =5730 years

 SOLUTION:

 First we will calculate fraction of sample

         x=A₁/A₂ =0.109/0.296=0.3682

Now,we will calculate the number of half lives

x = (1/2)ⁿ 

0.3682 = (1/2)ⁿ 

Taking Log O.B.S

log⁡ 0.3682=log⁡(1/2)ⁿ 

Now using property

       log⁡ aⁿ = n log a

we get

log⁡ 0.3682= n log⁡(1/2)

  -  0.4339  = n(-0.3010)

n= (-0.4339)/(-0.3010)=1.441

Now, the age of sample is given by

       t=n × T_1/2 =1.441×5730 = 8258 years

RESULT: The age of sample is 8258 years.