By Sunil Bhardwaj

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The disintegration of radio active element leads to the formation of a daughter element. If daughter element is radioactive it will also undergo the disintegration to produce another daughter element. The process goes on till a stable non-radio active element is formed.

$$A \xrightarrow [ of\ B ]{ Rate\ of\ Formation } B \xrightarrow [ of\ B ]{ Rate\ of\ Disintegration } C$$That state where the rate of formation of B from A is the same as its rate of disintegration into C, the amount of B in the source will remain unaltered. It will not change with time. Then B is said to have attained Radio active equilibrium. It is irreversible. The condition for radio active equilibrium is that the rate of formation of a daughter element must be equal to its rate of disintegration.

Rate of formation of B = Rate of disintegration of B.

But, Rate of disintegration of A = Rate of formation of B

$$\therefore$$ Rate of disintegration of A = Rate of disintegration of B $$\therefore -\frac { d{ N }_{ A } }{ dt } = -\frac { d{ N }_{ B } }{ dt }$$ But, $$\frac { -d{ N } }{ dt } = \lambda N$$ $$\therefore { \lambda }_{ A }{ N }_{ A } = { \lambda }_{ B }{ N }_{ B }$$ or $$\frac { { N }_{ A } }{ { N }_{ B } } = \frac { { \lambda }_{ B } }{ { \lambda }_{ A } }$$ also $$\lambda = \frac { 0.693 }{ { t }_{ { 1 }/{ 2 } } }$$ $$\therefore \boxed { \frac { { N }_{ A } }{ { N }_{ B } } = \frac { { \lambda }_{ B } }{ { \lambda }_{ A } } = \frac { { t }_{ { 1 }/{ 2 }A } }{ { { t }_{ { 1 }/{ 2 }B } } } }$$

Applications:

1. If we know the amount of two radio elements and if the half-life of one is known, we can find out of the other (OR) vice versa.

2. If we know the amount of two and decay constant of any one of it, we can find out that of the other and vice versa.

3. The radio active equilibrium mentioned above is known as Secular Equilibrium. The decay constants of the elements in the equilibrium say $${ \lambda }_{ A }$$ and $${ \lambda }_{ B }$$, differ from each other, widely. But if $${ \lambda }_{ A }$$ and $${ \lambda }_{ B }$$ not differing much from each other, then, This type of equilibrium is known as Transient (Temporary) Equilibrium.

A uranium mineral contains radium having reached secular equilibrium with uranium. The half life of radium is 1600 years. The mineral, on analysis, shows that it contains $$2.812 \times { 10 }^{ 6 }$$ atoms of uranium for each atom of radium. What is the half life of uranium?

We have,

Number of atoms of $$U = { N }_{ U } = 2.812 \times { 10 }^{ 6 }$$

Number of atoms of $$Ra = { N }_{ Ra } = 1$$

Half Life of $$Ra = { \left( { t }_{ 1/2 } \right) }_{ Ra } = 1600 years$$

Half Life of $$U = { \left( { t }_{ 1/2 } \right) }_{ U } = ?$$

According to radioactive equilibrium, $$\boxed { \frac { { N }_{ U } }{ { N }_{ Ra } } = \frac { { \left( { t }_{ 1/2 } \right) }_{ U } }{ { \left( { t }_{ 1/2 } \right) }_{ Ra } } } \qquad ...(1)$$ Lets Put the values, $$\frac { 2.812 \times { 10 }^{ 6 } }{ 1 } = \frac { { \left( { t }_{ 1/2 } \right) }_{ U } }{ 1600 }$$ $$\therefore { \left( { t }_{ 1/2 } \right) }_{ U } = 2.812 \times { 10 }^{ 6 } \times 1600$$ $$\therefore { \left( { t }_{ 1/2 } \right) }_{ U } = 4499.2 \times { 10 }^{ 6 }$$ $$= 4.499 \times { 10 }^{ 9 } years$$

One gram of radium is in radioactive equilibrium with $$0.59 { mm }^{ 3 }$$ of radon at NTP. The half life of radium is 1600 years. What is the half life (in days) of radon? (Given that: At. wt. of Ra = 226, At. wt. of Rn = 222)

We have,

Weight of $$Ra = 1 gm = \frac { 1 }{ 226 } moles = 4.425 \times { 10 }^{ -3 } moles$$

Volume of $$Rn = 0.59 { mm }^{ 3 } = 0.59 \times { 10 }^{ -3 } { cm }^{ 3 }$$

as per the relation, $$22,400 { cm }^{ 3 } of \ ideal \ gas \ at \ NTP \equiv 1 \ mole \ of \ ideal \ gas$$ $$\therefore 0.59 \times { 10 }^{ -4 } { cm }^{ 3 } of \ Rn \ gas \ at \ NTP \equiv x \ mole \ of \ Rn \ gas$$ $$\therefore x = \frac { 0.59 \times { 10 }^{ -3 } { cm }^{ 3 } }{ 22,400 { cm }^{ 3 } } = 2.634 \times { 10 }^{ -8 } moles$$

Half Life of $$Ra = { \left( { t }_{ 1/2 } \right) }_{ Ra } = 1600 years$$

Half Life of $$Rn = { \left( { t }_{ 1/2 } \right) }_{ Rn } = ?$$

According to radioactive equilibrium, $$\boxed { \frac { { N }_{ Rn } }{ { N }_{ Ra } } = \frac { { \left( { t }_{ 1/2 } \right) }_{ Rn } }{ { \left( { t }_{ 1/2 } \right) }_{ Ra } } } \qquad ...(1)$$ Lets Put the values, $$\frac { 2.634 \times { 10 }^{ -8 } }{ 4.425 \times { 10 }^{ -3 } } = \frac { { \left( { t }_{ 1/2 } \right) }_{ U } }{ 1600 }$$ $$\therefore { \left( { t }_{ 1/2 } \right) }_{ U } = \frac { 2.634 \times { 10 }^{ -8 } }{ 4.425 \times { 10 }^{ -3 } } \times 1600$$ $$= 9.524 \times { 10 }^{ -3 } years$$ $$= 9.524 \times { 10 }^{ -3 } \times 365 days$$ $$= 3.476 days$$

$$^{ 235 }{ U }$$ has a half life of $$7.2 \times { 10 }^{ 8 }$$ years and its daughter element $$^{ 231 }Th$$ has a half life of 24.75 hours. What is the weight of $$^{ 231 }Th$$ in equilibrium with 1 gm of $$^{ 235 }{ U }$$.

We have,

Half Life of $$^{ 235 }{ U } = { \left( { t }_{ 1/2 } \right) }_{ U } = 7.2 \times { 10 }^{ 8 }$$years

= $$7.2 \times { 10 }^{ 8 } \times 365days \times 24hours$$

= $$6.31 \times { 10 }^{ 12 } hours$$

Half Life of $$^{ 231 }Th = { \left( { t }_{ 1/2 } \right) }_{ Th } = 24.75 hours$$

Weight of $$^{ 235 }{ U } = 1 gm$$

Weight of $$^{ 231 }Th = ?$$

According to radioactive equilibrium, $$\boxed { \frac { { N }_{ Th } }{ { N }_{ U } } = \frac { { \left( { t }_{ 1/2 } \right) }_{ Th } }{ { \left( { t }_{ 1/2 } \right) }_{ U } } } \qquad ...(1)$$ also, $${ N }_{ Th } = \frac { Weight \ of \ Thorium }{ Atomic \ Wt \ of \ Thorium } \times Avogadros \ No.$$ $${ N }_{ U } = \frac { Weight \ of \ Uranium }{ Atomic \ Wt \ of \ Uranium } \times Avogadros \ No.$$ Lets put these in equation (1) $$\frac { { N }_{ Th } }{ { N }_{ U } } = \frac { { W }_{ Th } }{ { W }_{ U } } \frac { { (At. Wt) }_{ U } }{ { (At. Wt) }_{ Th } } = \frac { { \left( { t }_{ 1/2 } \right) }_{ Th } }{ { \left( { t }_{ 1/2 } \right) }_{ U } } \qquad ...(2)$$ Lets Put the values, $$\frac { { W }_{ Th } }{ 1 gm } \frac { 235 }{ 231 } = \frac { 24.75 }{ 6.31 \times { 10 }^{ 12 } }$$ $$\therefore W_{ Th } = \frac { 24.75 }{ 6.31 \times { 10 }^{ 12 } } \times \frac { 231 }{ 235 }$$ $$= 3.856 \times { 10 }^{ -12 } gm$$

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