### Radiation measurement of a radioactive sample

The rate at which a radioactive sample disintegrates can determine by counting the number of particles emitted in a given time.Radioactivity is one important natural phenomenon obeying the first-order kinetics. The rate of these reactions depends only on the single power of the concentration of the reactant.

(dN/dt) = - k N

where N = number of the atoms of the disintegrating radio-element, dt = time over which the disintegration is measured, and k = rate constant.

k = - (dN/dt)/N

The rate constant defined as the fraction decomposing in the unit time interval provided the concentration of the reactant 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 the t time interval. Rearranging and integrating over the limits N₀ and N and time, 0 and t.

Radiation measurement |

#### How to calculate the half-life of radioactive elements?

After a certain period of time, the value of (N₀/N ) becomes one-half and half of the radioactive elements have undergone disintegration. This period is called half-life of a radioactive element.Half-life of radio-element |

2.303 log(N₀/N ) = kt

when t = t

∴ 2.303 log{N₀/(N₀/2)} = k t

or, t

∴ k = 0.693/t

when t = t

_{½}, N = N₀/2.∴ 2.303 log{N₀/(N₀/2)} = k t

_{½}or, t

_{½}= 0.693/k∴ k = 0.693/t

_{½}.This relation shows that both the half-life and radioactive decay rate constants are independent of the amount of the radio-element present at a given time.

t

_{½}for polonium - 213 = 4.2 × 10⁻⁶ sec and bismuth - 209 = 3 × 10⁷ years.

### The average life period of radio-elements

The 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 the decay of all the atoms at the same time.

t

_{av}= total life/total number of atomsLet 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 lifetime of all the dN atoms is t dN.

Again the total number of atoms N₀ is composed of many such small numbers of atoms dN₁, dN₂, dN₃, etc, each with its own life span t₁, t₂, t₃, etc.

The average life of radio-element is reciprocal of its radioactive disintegration constant. This result can also be derived in a very simple way.

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.

t

or, t

∴ t

_{av}k = 1or, t

_{av}= 1/k∴ t

_{½}= 0.693/k = 0.693 t_{av}#### Radiocarbon dating - age of organic material

Radiocarbon dating is a method for determining the age of organic martial based on the accurate determination of the ratio of isotopes of carbon.The radiocarbon dating method was developed by Willard Libby, the University of Chicago in 1940 and receive a Nobel prize in chemistry for his work in 1960.

Radiocarbon - 14 is produced in the atmosphere by the interaction of neutron with ordinary nitrogen or cosmic reaction.

₇N¹⁴ + ₁n⁰ → ₆C¹⁴ + ₁H¹

Carbon reacts with atmospheric oxygen to form carbon dioxide. This carbon dioxide is taken by plants by photosynthesis and animals by eating plants.

When the animal or plant dies, it stops exchanging carbon with its environment since there no fresh intake of stratospheric carbon dioxide and the dead matter is out of equilibrium with the atmosphere.

The radiocarbon - 14 continues to decay so that thereafter a number of years only a fraction of carbon - 14 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(t½ = 5760 years)

By measuring this ratio and comparing it with the ratio in living plants one can estimate when the plant died.

Problem

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

Solution

We know that radioactive decay constant,

k = 0.693/(t½) = 0.693/5760 years

= 1.20 ×10⁻⁴ yr⁻¹

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

k = 0.693/(t½) = 0.693/5760 years

= 1.20 ×10⁻⁴ yr⁻¹

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

#### Age of rock deposits by half-life

Knowledge of the rate of decay of certain radioactive isotopes helps to determine the age of various rock deposits.Let us consider uranium-containing rock formed many years ago. The uranium started to decay giving rise to the uranium - 235 to lead -207 series.

The half-life of the intermediate members being small compared to that of uranium -235 (4.5 × 10⁹ years). Uranium atoms that started decaying many-many years ago must have been completely converted to the stable lead-207 during this extra-long period.

The uranium-235 remaining and the lead-207 formed must together account for the uranium 235 present at zero time when the rock solidified.

Thus both N₀ and N are known k is known from the knowledge of the half-life of uranium -235. Therefore the age of the rock can be calculated.

Problem

A sample of uranium (t₁/t₂ = 4.5 × 10⁹) ore is found to contain 11.9 gm of uranium-235 and 10.3 gm of lead-207. Calculate the age of the ore.

Solution

11.9 gm of uranium-238 = 11.9/238

= 0.05 mole of uranium

10.3 gm of lead-206 = 10.3/206

= 0.05 mole of lead -206

Mole of uranium -238 present in the ore at zero tim

= (0.05+0.05)

= 0.010 mole.

∴ Radioactive decay constant = 0.693/(4.5 × 10⁹)

= 0.154 × 10⁻⁹ yr⁻¹.

= 0.05 mole of uranium

10.3 gm of lead-206 = 10.3/206

= 0.05 mole of lead -206

Mole of uranium -238 present in the ore at zero tim

= (0.05+0.05)

= 0.010 mole.

∴ Radioactive decay constant = 0.693/(4.5 × 10⁹)

= 0.154 × 10⁻⁹ yr⁻¹.

2.303 log(0.10/0.05) = kt

∴ t = (2.303 log2)/(0.154 × 10⁻⁹ yr⁻¹)

= 4.5 × 10⁹ year

#### Avogadro number of radium - 226

Let 1 gm of radium - 226 contains N number of atoms.

∴ N = N₀/226

where N₀ = Avogadro number and mass number of radium = 226.

kN = (k × N₀)/m

where k = 0.693/t

kN = (0.693 × N₀)/(t

∴ N₀ = (kN × t

One gram of radium - 226 undergoes 3.7 × 10¹⁰ disintegrations per second and half-life = 1590 year.∴ N = N₀/226

where N₀ = Avogadro number and mass number of radium = 226.

kN = (k × N₀)/m

where k = 0.693/t

_{½}kN = (0.693 × N₀)/(t

_{½}× m)∴ N₀ = (kN × t

_{½}× m)/0.693∴ Avogadro number (N₀) of radium - 226

= (kN × t½ × m)/0.693

= (3.7 × 10¹⁰ × 1590 × 365 × 24 × 60 × 226)/0.693

= 6.0 × 10²³.

= (kN × t½ × m)/0.693

= (3.7 × 10¹⁰ × 1590 × 365 × 24 × 60 × 226)/0.693

= 6.0 × 10²³.