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The disintegration constant, half-life and average life of radio element are known as radio-active constants. The three radio active constants are related to one another and hence if any one is known the other two can be calculated. Generally, experimentally half-life period can be determined. The following are the methods of determination of half-life period of radio-elements.
1. For radio element with moderately short half-life.
When the radio elements have half-lives between a fraction of a second to several months, this method can be used. Here the rate of disintegration is easily measurable by means of G.M.counter (or) scintillation counters and this rate is proportional to the no. of unstable nuclei present.
We know, radioactive decay constant \lambda is given by, $$ \lambda = \frac { 2.203 }{ t } \log { \frac { { N }_{ 0 } }{ { N }_{ t } } } $$ where t is time,
\({ N }_{ 0 }\) is number of nuclei present initially and
\({ N }_{ t }\) is number of nuclei present after time t
$$ \therefore \frac { \lambda t }{ 2.303 } = \log { \frac { { N }_{ 0 } }{ { N }_{ t } } } $$ $$ \log { { N }_{ 0 } } - \log { { N }_{ t } } = \frac { \lambda t }{ 2.303 } $$ $$ or \log { { N }_{ t } } = \log { { N }_{ 0 } } - \frac { \lambda t }{ 2.303 } $$ When \(\log { { N }_{ t } } \) V/s t is plotted, the graph is straight line with negative slope. The slope is \(\frac { \lambda }{ 2.303 } \)
Hence, \(\lambda = 2.303 \times slope\).
Once \(\lambda\) is known to us then by using $${ t }_{ { 1 }/{ 2 } } = \frac { 0.693 }{ \lambda } $$ we can find out \({ t }_{ { 1 }/{ 2 } }\).
2. For radio element with moderately long half-lives.
In this case the rate of disintegration is very slow and the time required for disintegration may be years. Therefore no. of \(\alpha \) or \(\beta \) particles emitted \((d{ N }_{ t })\) in a definite time interval \( (dt)\) an hour or a day is measured experimentally and then substituted in the expression, $$ \frac { -d{ N }_{ t } }{ dt } = \lambda { N }_{ t } $$ $$ or \lambda = \frac { -d{ N }_{ t } }{ { N }_{ t }dt } $$ Where \({ N }_{ t }\) is the no. of nuclei of radio-element, which is calculated as, $$ { N }_{ t } = \frac { Wt. \ of \ radio \ element \ (W) }{ Atomic \ mass \ (A) } \times Avogadros \ Number $$ So by knowing all these values we can easily find \(\lambda \), then from \(\lambda\) we can find out $${ t }_{ { 1 }/{ 2 } } = \frac { 0.693 }{ \lambda } $$
3. For radio elements with very short (OR) very long half-lives.
Here we have to apply the principle of Radio active Equilibrium (OR) Geiger - Nuttal relationship.
(a) From radio active equilibrium:
We know that,$$ \frac { { N }_{ A } }{ { N }_{ B } } = \frac { { \lambda }_{ B } }{ { \lambda }_{ A } } = \frac { { t }_{ { 1 }/{ 2 }A } }{ { { t }_{ { 1 }/{ 2 }B } } } $$ In this case \({ N }_{ A }\) and \({ N }_{ B }\) are determined by analysis if decay constant of any one the radio element is known, we can find out of the other.
(b) Geiger-Nuttal Rule:
Geiger - Nuttal equation is, $$ \log { \lambda } = A \log { R } + B $$ for \(\alpha\) -particles in air:
R = Range of \alpha -particles in air.
A = Constant which is nearly same for all series.
B = constant but changes from radio-active series to series.
A, B and R are determined experimentally in air then by using Geiger - Nuttal equation, we can find out \(\lambda \) and if \(\lambda \) is known, we can find out $${ t }_{ { 1 }/{ 2 } } = \frac { 0.693 }{ \lambda } .$$
Alpha particles from RaC source where found to have a range of 3.0 cm. Using Geiger-Nuttal rule calculate the decay constant and hence the half life of RaC in days. (Give A = 61.5 and B = -38.5 with half life in days)
We have, Range R = 3.0 cm According to Geiger Nuttal rule, $$ \boxed { \log _{ 10 }{ \lambda } = A \log _{ 10 }{ R } + B } \qquad ...(1) $$ Lets put these in equation (1) $$ \log _{ 10 }{ \lambda } = \left( 61.5 \right) \log _{ 10 }{ \left( 3.0 \right) } + \left( -38.5 \right) $$ $$ \log _{ 10 }{ \lambda } = 61.5 \times \left( 0.477 \right) + \left( -38.5 \right) $$ $$ \log _{ 10 }{ \lambda } = 29.343 - 38.5 $$ $$ \log _{ 10 }{ \lambda } = -9.157 $$ $$ \therefore \lambda = A.\log { \left( -9.157 \right) } = 6.966 \times { 10 }^{ -10 } { days }^{ -1 } $$ Now $$ { t }_{ 1/2 } = \frac { 0.693 }{ \lambda } $$ $$= \frac { 0.693 }{ 6.966 \times { 10 }^{ -10 } } $$ $$= 0.995 \times { 10 }^{ 9 } days $$
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