question_answer

Explain what is meant by radioactive decay. A radioactive nucleus is represented by the symbol \[_{b}^{a}V\]. How is the new nucleus represented after the emission of (i) an alpha particle, (ii) a beta particle and (iii) a gamma ray? The activity of a source under- going a single type of decay is \[{{R}_{0}}\] at time \[t=0\]. Obtain an expression in terms of the half-life. \[{{T}_{1/2}}\] for the activity R at any subsequent time t.                                             

Answer:

                Radioactive decay is the spontaneous disintegration of the nucleus of an atom with the emission of one or more penetrating radiations like \[\alpha ,\beta \]and \[\gamma \]-rays. (i) \[\alpha \]-decay: \[_{b}^{a}V\to _{b}^{a}_{-2}^{-4}X+_{2}^{4}He+Q\] (ii) \[\beta \] decay: \[_{b}^{a}V\to _{b+1}^{a}X+_{-1}^{0}e+\overset{-}{\mathop{v}}\,+Q\] (iii) \[\gamma \]-decay: \[\underset{\text{state}}{\mathop{\underset{\text{Excited}}{\mathop{_{b}^{a}V}}\,}}\,\to \underset{\text{state}}{\mathop{\underset{\text{Ground}}{\mathop{_{b}^{a}V+\gamma }}\,}}\,\] Activity, \[R=-\frac{dN}{dt}\] According to radioactive decay law,  \[-\frac{dN}{dt}=\lambda N\] \[\therefore \] \[R=\lambda N\] As           \[N={{N}_{0}}{{e}^{-\lambda t}}\] \[\therefore \]  \[R=\lambda N{{e}^{-\lambda t}}\] or \[R={{R}_{0}}{{e}^{-\lambda t}}\] where \[\lambda {{N}_{0}}={{R}_{0}}=\] activity of the sample at \[t=0\]. Also       \[\lambda =\frac{0.693}{{{T}_{1/2}}}\] \[-\frac{0.693t}{{{T}_{1/2}}}\] \[\therefore \]  \[R={{R}_{0}}e\]