### Rate of radioactive decay half life

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. The rate of these reactions depends only on the single power of the concentration. of the reactant. The rate can be expressed as,

(dN/dt) = - k N

- Where N is the number of the atoms of the disintegrating radio-element present at any time, dt is the time over which the disintegration is measured and k is the radioactive decay rate constant.

k = - (dN/dt)/N

- Thus, the rate constant(k) is defined as the fraction decomposing in the unit time interval provided the concentration of the reactant is 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.

Rate of radioactive decay half-life |

#### Half-life period of radioactive elements

- 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 a radioactive element and is a characteristic property of a radioactive element.

Half-life and rate of radioactive decay |

- If

**radioactivity**of an element is 100% and the half-life period of this element 4 hours. Thus after four hours, it decomposes 50% and the remaining 50%. After 8 hours it decomposes 75% and reaming 25% and the process is continued. The half-life is given by,

2.303 log(N₀/N ) = kt When t = t

_{½}, N = N₀/2Putting in the above equation we have,

2.303 log{N₀/(N₀/2)} = k t

or, t

2.303 log{N₀/(N₀/2)} = k t

_{½}or, t

_{½}= 0.693/kk = 0.693/t

_{½}- The above 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.

- ₈₄Po²¹³ has t

_{½}= 4.2 × 10⁻⁶ sec, whereas ₈₃Bi²⁰⁹ is 3 × 10⁷ years.

### Age of radioactive elements

#### 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 thereafter 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(t½ = 5760 years)

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

#### Age of rock deposits

- 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 - 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.

#### Avogadro number

If 1 gm of a radioactive element contains N number of atoms.

Then, N = N₀/m

where N₀ = Avogadro number and m = Mass number

Therefore, kN = (k × N₀)/m

where k = 0.693/t

Thus, kN = (0.693 × N₀)/(t

Then, N = N₀/m

where N₀ = Avogadro number and m = Mass number

Therefore, kN = (k × N₀)/m

where k = 0.693/t

_{½}Thus, kN = (0.693 × N₀)/(t

_{½}× m)∴ N₀ = (kN × t

_{½}× 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 disintegrations per second, the mass number of Ra-226 is 226 and t

_{½}= 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

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

= 6.0 × 10²³

### Average life period of radioactive elements

- 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 aggregate of a large number of atoms.

- 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. The average life period may be calculated as follows-

t

_{av}= 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 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.

Average life and the rate of radioactive decay |

- 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.

t

_{av}k = 1 So that, t_{av}= 1/k- Thus the relation between an average life and half-life is,

t

Problem_{½}= 0.693/k = 0.693 t_{av}- 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).

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

Problemk = 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

- A sample of uranium (t₁/t₂ = 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

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.

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

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

= 4.5 × 10⁹ year