By Sunil Bhardwaj

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The decay of the nucleus in the certain time interval does not depend on its chemical state and is independent of temperature, and pressure. The rate of radioactive decay depends only on number of nuclei present. As the rate depends only on number of nuclei present, therefore we can say, the rate of decay is FIRST ORDER REACTION.

If dN are the no. of the atoms disintegrated in dt time.
\( \therefore \frac { dN }{ dt } \) is it is rate of disintegration.
$$ \therefore \frac { dN }{ dt } \propto N $$ where N is the no. of atoms present at that instant.
As the rate of decay goes on decreasing we can add a negative sign.$$ \therefore -\frac { dN }{ dt } \propto N $$ $$ or -\frac { dN }{ dt } = \lambda N $$ where \(\lambda \) is decay constant. $$ \therefore \boxed { -\frac { 1 }{ N } dN = \lambda dt } \qquad ....(1)$$ so decay constant can be defined as the fraction of the total no. of atoms which are distintegrated in a unit time.
On integrating the above equation between limits: \(N = { N }_{ 0 } \) to \( N = { N }_{ t }\) and t = 0 to t = t. $$ \therefore \int _{ { N }_{ 0 } }^{ { N }_{ t } }{ -\frac { 1 }{ N } dN } = \lambda \int _{ 0 }^{ t }{ dt } $$ $$ \therefore { -\left[ \ln { N } \right] }_{ { N }_{ 0 } }^{ { N }_{ t } } = \lambda { \left[ t \right] }_{ 0 }^{ t }$$ $$ \therefore -\left[ \ln { { N }_{ t } } - \ln { { N }_{ 0 } } \right] = \lambda \left[ t - 0 \right] $$ $$ \therefore -\ln { \frac { { N }_{ t } }{ { N }_{ 0 } } } = \lambda t$$ $$ \therefore \ln { \frac { { N }_{ t } }{ { N }_{ 0 } } } = -\lambda t$$ $$ \therefore \frac { { N }_{ t } }{ { N }_{ 0 } } = { e }^{ -\lambda t }$$ $$ \therefore \boxed { { N }_{ t } = { N }_{ 0 }{ e }^{ -\lambda t } } \qquad ....(2)$$ From this equation it is clear that \({ N }_{ t }\) decreasing exponentially with time t.

The half life is defined as the time in which half of the initial amount of radio element is disintegrated.

i.e. when \({ N }_{ t } = \frac { { N }_{ 0 } }{ 2 }\) equation (2) becomes, $$ \frac { { N }_{ 0 } }{ 2 } = { N }_{ 0 }{ e }^{ -\lambda t } $$ $$ \therefore { e }^{ -\lambda t } = \frac { 1 }{ 2 } = 0.5 $$ $$ \therefore -\lambda t = \ln { \left( 0.5 \right) } = -0.693 $$ $$ \therefore t = \frac { 0.693 }{ \lambda } $$ $$ or \ \boxed { { t }_{ { 1 }/{ 2 } } = \frac { 0.693 }{ \lambda } } \qquad ...(3) $$ Thus half life is Inversely proportional to decay constant. and is independent of initial amount of radio element. If either the decay constant \((\lambda )\) or half life \(({ t }_{ { 1 }/{ 2 } })\) is known, other constants for radio element can be determined.

Numerical:The half life of \(^{ 14 }{ C }\) is 5600 years. Calculate the rate of disintegration per minute for 1 gm of it. (Avogadro’s Number = \(6.023 \times { 10 }^{ 23 }\))

Solution: We have,$$ { t }_{ 1/2 } = 5600 years, $$ $$ \qquad = 5600 \times 365 days \times 24 Hrs \times 60 Min $$ $$ \qquad = 2.94336 \times { 10 }^{ 9 } min $$ Rate of disintegration is given by, $$ \boxed { -\frac { dN }{ dt } = \lambda { N }_{ t } } $$ So lets find \(\lambda \) from the relation $$ { t }_{ 1/2 } = \frac { 0.693 }{ \lambda } $$ $$ \therefore \lambda = \frac { 0.693 }{ { t }_{ 1/2 } } $$ $$= \frac { 0.693 }{ 2.94336 \times { 10 }^{ 9 } } $$ $$= 2.35445 \times { 10 }^{ -10 } min^{ -1 } $$ and \({ N }_{ t }\) can be calculated from, $$ { N }_{ t } = \frac { Wt \ of \ radio \ element }{ At. \ mass \ of \ radio \ element } \times \ Avogadro’s \ Number $$ $$ = \frac { 1 gm }{ 14 } \times \left( 6.023 \times { 10 }^{ 23 } \right) $$ $$ = 4.302 \times { 10 }^{ 22 }gm $$ Let’s put these values in equation (1) $$ -\frac { dN }{ dt } = \left( 2.35445 \times { 10 }^{ -10 } min^{ -1 } \right) \times \left( 4.302 \times { 10 }^{ 22 }gm \right) $$ $$ = 1.013 \times { 10 }^{ 13 }gm/min$$

Numerical:Half life of a radioelement if 5 minutes. What fraction of it will remain after 25 minutes.Solution:We have, $$ time t = 25 minutes $$ $$ { t }_{ 1/2 } = 5 minutes $$ $$ \therefore \lambda = \frac { 0.693 }{ { t }_{ 1/2 } } = \frac { 0.693 }{ 5 } = 0.1386 $$ Fraction remaining can be calculated using, $$ \boxed { \lambda = \frac { 2.303 }{ t } \log _{ 10 }{ \frac { { N }_{ 0 } }{ { N }_{ t } } } } \qquad ...(1) $$ Lets Put the values, $$ 0.1386 = \frac { 2.303 }{ 25 } \log _{ 10 }{ \frac { { N }_{ 0 } }{ { N }_{ t } } } $$ $$ \log _{ 10 }{ \frac { { N }_{ 0 } }{ { N }_{ t } } } = \frac { 0.1386 \times 25 }{ 2.303 } = 1.505 $$ $$ \frac { { N }_{ 0 } }{ { N }_{ t } } = { 10 }^{ 1.505 } = 31.957 $$ fraction remaining after 25 minutes, $$ \frac { { N }_{ t } }{ { N }_{ 0 } } = \frac { 1 }{ 31.957 } = 0.0313 $$