Radioactive Equilibrium





Radioactive Equilibrium

In radioactive transformation when a daughter element transforms at the same rate at which it is formed from the parent element that state is known as radioactive equilibrium e.g.

A \; \; \longrightarrow \; \; B \; \; \longrightarrow \; \; C

parent    intermediate daughter

At equilibrium

\lambda_A N_A = \lambda_B N_B

Or, \dfrac{\lambda_A}{\lambda_B} = \dfrac{N_B}{N_A} = \dfrac{t_{1/2(B)}}{t_{1/2(A)}} \\ \dfrac{N_B}{N_A} = \dfrac{\lambda_A}{\lambda_B- \lambda_A}

 

Example 1: ^{232}_{90} Th disintegrates to give ^{206}_{82} Pb as the final product. How many alpha and beta particles are emitted during this process?

Solution: in this process change in at. Wt. =234 – 206 = 28 a.m.u. since by the emission of one \alpha-particle m. no. decreases by 4 units hence no. of emitted \alpha-particle = 28/4 = 7

If only \alpha-particles emit then the at. No. of the product should be 90 – 2 × 7 = 76.

But here the at. no. of the product is 82 that is due to the eimission of \beta-particles and its no. is 82 – 76 = 6

 

Example 2: The activity of a radio isotope falls to 12.5% in 90 days. Compute the half life and decay constant of the radio isotope.

Solution:

Since,

\lambda = \dfrac{2.303}{t} \log \dfrac{N_0}{N} = \dfrac{2.303}{90 \text{days}} \log \dfrac{100}{125} \\[3mm] \lambda = 2.308 \times 10^{-2} day^{-1} \\[3mm] \text{now}, t_{1/2} = \dfrac{0.693}{\lambda} = \dfrac{0.693}{2.308 \times 10^{-2}} \text{days} \\[3mm] 30 \text{days}

 

Example 3: Calculate the weight in gram of 1 curie and 1rd of ^{214}_{82} Pb(t_{1/2} = 26.8 min)

Solution:

From eq. 3

\lambda = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{26 \times 60 s} = 4.31 \times 10^{-4} s^{-1}

Total no. of atoms in w gram ^{214}_{82}Pb = \dfrac{w}{214} \times 6.02 \times 10^{23}

 

From eq. 1,

 

-\dfrac{dN}{dt} = \lambda N = 4.31 \times 10^{-4} s^{-1} \times \dfrac{2}{214} \times 6.02 \times 10^{23} \\ = 1.21 \times 10^{18} w \text{disintegration} s^{-1}

 

According to definition of curie

 

3.7 \times 10^{10} g = 1.21 \times 10^{18} w \\ or \\ 2 = 3.1 \times 10^{-8} g

 

According to definition of Rutherford

 

10^6 g = 1.21 \times 10^{18} w \\ or \\ w = 8.3 \times 10^{-13} g

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