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__Rate of Radioactive Decay:
__

__Rate of Radioactive Decay:__

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**.
(dN/N) = - kdt

Mathematical form of Radioactive Reaction |

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):__

__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:__

__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.Plot for Radioactive Decay |

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 = t₁/₂, N = N₀/2

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.

₈₄

**Po**²¹³ has t₁/₂ = 4.2 × 10⁻⁶ sec, whereas ₈₃**Bi**²⁰⁹ is 3₀ × 10⁷ years.###
__Calculation of the Age Of Radioactive Elements__:

__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.

₇

**N**¹⁴ + ₀**n**¹ → ₆**C**¹⁴ + ₁**H**¹(__)__**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.

₆

**C**¹⁴ → ₇**N**¹⁴ + ₋₁**e**⁰ (**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:**

**(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:__

__Determination of the Avogadro Number:__

If

**1**gm of a radioactive element contain**N**number of atoms.
Then,

**N**= Avogadro Number/Mass Number
Therefore,

**k****N**= (**k**× Avogadro Number)/Mass Number
Where

**k**= Radioactive Decay Constant and**k**= 0.693/t₁/₂.
Thus,

**k**N = (0.693 × Avogadro Number)/(t₁/₂ × Mass Number)**∴ Avogadro Number = (kN × t₁/₂ × Mass Number)/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,

__Avogadro Number__= (3.7 × 10¹⁰ sec⁻¹ × 1590 × 365 × 24 × 60 sec × 226)/0.693

=

**6.0 × 10²³**###
__The Average Life Period(tav):__

__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**