Directions: In this question, a word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in the two matrices, given below. The columns and rows of Matrix (I) are numbered from 0 to 3 and those of Matrix (II) are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, e.g. E can be represented by 00,11, etc. and I can be represented by 56, 86, etc. Similarly, you have to identify the set for the word 'TOLD'.

67, 30, 55, 13

76, 02, 13, 55

76, 33, 55, 31

85, 02, 23, 31

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Concepts Covered - 1

Simultaneous and Series Disintegration

Simultaneous decay-

As we know that due to radioactive disintegration a radio nuclide transforms into its daughter nucleus. Depending on the nuclear structure and its unstability, a parent nucleus may undergo either $\alpha-$ or $\beta-$ emission. Sometimes a parent nucleus may undergo both types of emission imultaneousonly.

If an element decays to different daughter nuclei with different decay constant $\lambda_{1},\;\lambda_{2},\;\lambda_{3},\ldots$ etc. for each decay mode, then the effective decay constant of the parent nuclei can be given as

$\lambda_{eff}=\lambda_{1}+\lambda_{2}+\lambda_{3},\ldots$

Similarly, for a radioactive element with decay constant $\lambda$ which decays by both $\alpha-$ and $\beta-$ decays given that the probability
for an $\alpha-$ -emission is $\mathrm{P_1}$ and that for $\beta-$ emission is $\mathrm{P_2}$ the decay constant of the element can be split for individual decay modes. Like in this case the decay constants for $\alpha-$ and $\beta-$ decay separately can be given as

$\begin{array}{l}{\lambda_{\alpha}=P_{1} \lambda} \\ {\lambda_{\beta}=P_{2} \lambda}\end{array}$

Series decay-

Accumulation of Radioactive element in Radioactive series-

A radioactive element decays into its daughter nuclei until a stable element appears. Consider a radioactive series-

$A_1\overset{\lambda_1}{\rightarrow}A_2\overset{\lambda_2}{\rightarrow}A_3\overset{\lambda_3}{\rightarrow} \dots$

A radioactive element $A_1$ disintegrates to form another radioactive element $A_2$ which in turn disintegrates to another element $A_3$ and so on. Such decays are called Series or Successive disintegration.

Here, rate of disintegration of $A_1$ = Rate of formation of B

$\frac{-dN_A_1}{dt}=\frac{dN_A_2}{dt} = \lambda_1 N_{A1}$

$\frac{-dN_A_2}{dt}=\frac{dN_A_3}{dt} = \lambda_2 N_{A2}$

$\frac{dN_A_1}{dt} = -\lambda_1 N_{A1}$

$\frac{dN_A_2}{dt} = -\lambda_2 N_{A2}$

Therefore, net formation of B = Rate of disintegration of $A_1$ - Rate of disintegration of $A_2$

= $\lambda_1 N_A_1 \ - \ \lambda_2 N_A_2$

If the Rate of disintegration of $A_1$ becomes equal to the Rate of disintegration of $A_2$ , then it is called Radioactive equilibrium. So the equation become -

$\Rightarrow \frac{\lambda_1}{\lambda_2} = \frac{N_A_2}{N_A_1} = \frac{T_{avg2}}{T_{avg1}} = \frac{(T_{\frac{1}{2}})_2}{(T_{\frac{1}{2}})_1}$