A Page about Chemistry and Related Topics.

## Oct 31, 2018

The rate at which a radioactive sample disintegrates can be determined by counting the number of particles emitted in a given time.
Radioactive decay is one important natural phenomenon obeying the first order rate process. Rate of these reactions depends only on the single power of the conc. of the reactant. The rate can be expressed as,
 (dN/dt) = - kN
Where N is the number of the atoms of the disintegrating radio-element present at the any time, dt is the time over which the disintegration is measured and k is the radioactive decay rate constant.
 Again, k = - (dN/dt)/N
Thus, the rate constant(k) is defined as the fraction decomposing in unit time interval provided the conc. of the reactant in kept constant by adding from outside during this time interval.
The negative sign shows that N decreases with time.
Let N0 = number of the atoms present at the time t=0 and N = Number of atom present after t time interval.
Rearranging and integrating over the limits N0 and N and time, 0 and t.
 N0 t ∫ dn/N = - ∫ k dt N 0
 N0 t Therefore, ∫ d(lnN) = - ∫ k dt N 0
Thus, ln(N/N0) = -kt
or, N/N0 = e-kt
Converting to the logarithm to the base 10,
we have,  2.303 log(N/N0) = - kt
or, 2.303 log(N0/N) = kt

### Evaluation of Rate Constant(k):

Using the above equation rate constant(k) can be determined by graphical plot
of log(N0/N ) against time t.
 Graphical Plot of log(N₀/N ) against time t

### Half Life Period of Radioactive Element:

After a certain period of time the value of (N0/N ) becomes one half, that is, half of the radioactive elements have undergone disintegration. This period is called half life of radioactive element and is a characteristic property of a radioactive element.
If a radioactive element is 100% radioactive and the half life period of this element 4 hour.
Thus after four hour it decompose 50% and remaining 50%. After 8 hour it decompose 75% and reaming 25% and the process is continued.
The half life is given by,
2.303 log(N0/N ) = kt
When t = t1/2N = N0/
Putting in the above equation we have,
2.303 log{N0/(N0/2)} = kt1/2
 or, t1/2 = 0.693/k  or, k = 0.693/t1/2
The above relation shows that both the half-life and radioactive decay rate constant are independent of the amount of the radio-element present at a given time.
84Po213 has t1/2 = 4.2 × 10-6 sec
whereas 83Bi209 is 3₀ × 107 years.

### Calculation of the Age Of Radioactive Elements:

(i) Age of Organic Material:
A method of determining the age of organic martial based on the accurate determination of the ratio of carbon-14 and carbon-12. carbon 14 is produced in the atmosphere by the interaction of neutron with ordinary nitrogen.
7N14 + 1n0  6C14 + 1H1
( Cosmic Reaction)
The carbon-14 ultimately goes over to carbon-14 dioxide. A steady state concentration of one 14C to 12C is reached in the atmospheric CO2 . This carbon dioxide is taken in or given out by plants and plant eating animals or human beings so they all bear this ratio.

When a plant or animals died the steady state is disturbed since there is no fresh intake of stratospheric CO2 the dead matter is out of equilibrium with the atmosphere. The 14C continues to decay so that there after a number of years only a fraction of it is left on the died matter. Therefore the ratio of the 14C/12C drops from the steady state ratio in the living matter.
6C14   7N14 + -1e0
(t1/2 = 5760 years)
By measuring this ratio and comparing it with the ratio in living plants one can estimate when the plant died.

A piece of wood was found to have ¹⁴C/¹²C ratio 0.7 times that in the living plant. Calculate the approximate period when the plant died(t₁/₂ = 5760 years).

 We know that radioactive decay constant,  k = 0.693/(t₁/₂)  = 0.693/5760 years  = 1.20 ×10⁻⁴ yr⁻¹,  and N₀/N = 1.00/0.70. ∴ 2.303 log(N₀/N) = kt or, t = (2.303 log(N₀/N)/k Putting the value, above equation, we have, t = (2.303 × 0.155)/(1.20 × 10⁻⁴)  = 2970 years

### (ii) Determination of the Age of Rock Deposits:

A knowledge of the rate of decay of certain radioactive isotopes helps to determine the age of various rock deposits. Let us consider a uranium containing rock formed many years ago.
The uranium started to decay giving rise to the uranium - 238 to lead -206 series. The half lives of the intermediate members being small compared to that of uranium -238(4.5 × 109 years), it is reasonable to assume that those uranium atoms that started decaying many many years ago must have been completely converted to the stable lead-206 during this extra long period.
The uranium-238 remaining and the lead-206 formed must together account for the uranium 238 present at zero time , that is, when the rock solidified. Thus both N0 and N are known k is known from a knowledge of the half life of uranium -238. Therefore the age of the rock can be calculated.

A sample of uranium (t₁/₂ = 4.5 × 10⁹) ore is found to contain 11.9 gm of uranium-238 and 10.3 gm of lead-206. Calculate the age of the ore.

11.9 gm of uranium-238 = 11.9/238 = 0.05 mole of uranium and 10.3 gm of lead-206 = 10.3/206 = 0.05 mole of lead -206.
Thus mole of uranium -238 present in the ore at zero time = (0.05+0.05) = 0.010 mole.
And radioactive decay constant = 0.693/(4.5 × 10⁹) = 0.154 × 10⁻⁹ yr⁻¹.
Then 2.303 log(0.10/0.05) = kt
∴ t = (2.303 log2)/(0.154 × 10⁻⁹ yr⁻¹)
= 4.5 × 10⁹ year.

### Determination of the Avogadro Number:

If 1 gm of a radioactive element contain N number of atoms.
Then, N = Avogadro Number(N0)/Mass Number(m
Therefore, kN = (k × N0)/m
Where k = Radioactive Decay Constant
= 0.693/t1/2
 Thus, kN = (0.693 × N0)/(t1/2 × m)  ∴ Avogadro Number(N0)  = (kN × t1/2 × m)/0.693
If each atom of the radio-element expels one particle then kN, the rate of decay, is also the rate at which such particles are ejected.
One gram of radium undergoes 3.7 × 1010 disintegration per second, mass number of Ra-226 = 226 and t1/2 = 1590 year, so that the Avogadro Number can be calculated by the above equation,
(3.7 × 1010 sec⁻¹ × 1590 × 365 × 24 × 60 sec × 226)/0.693
= 6.0 × 1023

### The Average Life Period(tav):

Besides half life of a radio-element another aspect of the life of a radio-element must also be known. It is now possible to determine the average life period of a radioactive atom present in an aggregate of a large number of atoms. Average life period of an atom of the radio-element tells us the average span of time after which the atom will disintegrate.

The length of time a radio-element atom can live before it disintegrates may have values from zero to infinity. This explains the gradual decay of the radio-element instead of decay of all the atoms at the same time.
The average life period may be calculated as follows-
 tav = Total Life Period/Total Number of Atoms
Let N0 atoms of a radioactive element are present in an aggregate of large no of atoms at time zero. Now in the small time interval t to (t+dt), dN atoms are found to disintegrate. Since dt is a small time period, we can take dN as the number of atoms disintegrating at the time t. So the total life time of all the dN atoms is tdN.
Again the total number of atoms N0 is composed of many such small number of atoms dN1, dN2, dN3 etc, each with its own life span t1, t2, t3 etc.
 Representation of Average Life Period of Radio-element
The average life of Radio-element is thus the reciprocal of its radioactive disintegration constant. This result can also be derived in a very simple alternative way. Since the radioactive atoms may be regarded as having an average life, then the product of the fractions of atoms disintegrating in unit time (that is k) and average life must be unity.
tavk = 1, So that tav = 1/k
Thus the relation between average life and half life is,
 t1/2 = 0.693/k = 0.693 tav