SATHEE: nuclei Question 37

nuclei Question 37

Question: Q. 12. (i) The number of nuclei of a given radioactive nucleus, at times $t = 0$ and $t = T$ , are $N_{0}$ and $(N_{0} / n)$ respectively. Obtain an expression for the half life $(T_{1 / 2})$ of this nucleus in terms of $n$ and $T$ .

(ii) Identify the nature of the ‘radioactive radiations’, emitted in each step of the ‘decay chain’ given below :

$^{A} X_{Z} \to^{A - 4} Y_{Z - 2} \to^{A - 4} Y_{Z - 2} \to^{A - 4} W_{Z - 1}$

U] [SQP 2013]

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Solution:

Ans. (i) According to the (exponential) law of radioactive decay,

Taking $\log$ on both sides

$\begin{matrix} (1) & \begin{aligned} \ln n & = - λ t & Now given & t & = T & and λ & = \frac{0.693}{T_{1 / 2}} & \frac{\ln n}{T} & = \frac{0.693}{T_{1 / 2}} ∴ & T_{1 / 2} & = \frac{0.693 T}{\ln n} \end{aligned} \end{matrix}$

(ii) In $^{A} X_{Z} \to^{A - 4} Y_{Z - 2}$ ; Mass number changes by 4 units and atomic number changes by 2 units. Hence, it is $α$ -decay. In ${ }^{A-4} \mathrm{Y}{\mathrm{Z}-2} \rightarrow{ }^{A-4} Y{\mathrm{Z}-2} $; t h e r e i s n o c h a n g e i n m a s s n u m b e r a n d a t o m i c n u m b e r . H e n c e i t i s$ \gamma$ - decay.

In $^{A - 4} Y_{Z - 2} \to^{A - 4} W_{Z - 1}$ ; there is no change in mass number and +1 change in atomic number. Hence this is $β^{-}$ decay.

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