Half-life

Evolution of a sample

Semi-desin- theft	Nucles without disintegrate
0	100% (all)
1	50% (1 in 2)
2	25% (1 in 4)
3	12.5 % (1 in 8)
4	6.25 % (1 in 16)
5	3.125 % (1 in 32)
6	1.562 5 % (1 in 64)
7	0.781 25% (1 in 128)
N	1002N% % {displaystyle {frac {100}{2^{N}}}{%} (1 in every 2N)

In nuclear physics and radiochemistry, the half-life or half-life constant, also called half-life or half-life, is defined as the time required to for half the nuclei in an initial sample of a radioisotope to decay. Half of them are taken as reference due to the random nature of nuclear decay.

The period of semi-disintegration should not be confused with the average life. This concept is widely used in the calculations of nuclear kinetics, in order to characterize the nuclides, as well as a pattern of nuclear purity of the samples. This constant is usually represented with Δ Δ {displaystyle tau }.

It can also be understood as the time it takes for half of the radioactive atoms in a sample to transmute. An example is carbon-14 used to date ancient organic remains.

Notation:

• t1/2{displaystyle t_{1/2} is the period of semi-deintegration.
• N(t){displaystyle N(t)} is the number of nuclei of the sample in the instant time t.
• N0{displaystyle N_{0}} is the initial number (when t = 0) of the sample cores.
• λ λ {displaystyle lambda } is the constant disintegration.

The moment the number of cores has been reduced by half is t1/2{displaystyle t_{1/2},}. I mean:

N(t1/2)=N0⋅ ⋅ 12{displaystyle N(t_{1/2})=N_{0}cdot {frac {1}{2}}{2}}}}

Substituting into the exponential decay formula:

N0⋅ ⋅ 12=N0e− − λ λ t1/2{displaystyle N_{0}cdot {frac {1}{2}}}=N_{0}e^{-lambda t_{1/2}}{,}

e− − λ λ t1/2=12{displaystyle e^{-lambda t_{1/2}}={frac {1{2}}}{,}

− − λ λ t1/2=ln 12=− − ln 2{displaystyle -lambda t_{1/2}=ln {frac {1}{2}}}=-ln {2},}

Therefore, the relationship between the semi-disintegration period of a radioisotope (t1/2{displaystyle t_{1/2}) and its constant disintegration (λ λ {displaystyle {lambda}}) is:

t1/2=ln 2λ λ {displaystyle t_{1/2}={frac {ln 2}{lambda },}

And like his average life (Δ Δ {displaystyle tau }It is

Δ Δ =1λ λ {displaystyle tau ={frac {1}{lambda }}}}

It turns out that the half-life is about 69.31% of its half-life.

If we want to calculate the time it takes for a sample of a radioisotope to reduce to 20% of the initial one, we will do:

Cor{displaystyle co} = Initial concentration.

Ct=0.2↓ ↓ Cor{displaystyle Ct=0.2*Co}

K{displaystyle K} = Semi-disintegration Constant

t1/2{displaystyle t_{1/2} = Semi-disintegration period

t1/2=ln (CorCt)k{displaystyle t_{1/2}={frac {ln({frac {Co}{Ct}}}}}}{k}}}}}

The rate of disintegration of a contaminant will be lower the less amount of contaminant remains (we assume that the contaminant follows first-order kinetics).

Half-lives of some radionuclides