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(a) Define the term ‘activity’ of a given sample of radio nuclide. Write the expression for the law of radioactive decay in terms of the activity of a given sample. (b) A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to 3.125% of its original value ? (c) When a nucleus (X) undergoes beta-decay, it transforms to the nucleus (Y). Does the pair (X, Y) form isotopes, isobars or isotones ? Justify your answer. |
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Answer» Solution :(a) Activity of a radio nuclide is defined as the rate of disintegration of the given sample of radio nuclide. Thus, activity `R=-(dN)/(dt)`. As per LAW of radioactive decay, activity of a radio nuclide is directly proportional to the QUANTITY (number of nuclei present) of given radio nuclide. Mathematically, `RpropN or R=lambdaN`,where `lambda` is a constant known as the disintegration constant and its value depends on the nature of radionuclide. As per above relation, the activity goes on DECREASING with time as per relation: `R_(t)=R_(0)e^(-lambdat)`, where `R_(0)`=activity at time t=0 and `R_(t)`=activity at time t. (b) Here `R_(t)=3.125% "of"R_(0)=3.125/100R_(0)=R_(0)/32=R_(0)(1/2)^(5)` As activity FALLS to one-half of its original value in one half-life. Hence, it will fall to `(1/2)^(5)R_(0)` in 5 half-lives. `therefore` Time `t=5T_(1/2)=5Tyears` (c) The `beta`-decay process can be expressed as: `" "_(Z)^(A)Xoverset(-beta^(-))(to)" "_(Z+1)^(A)Y+" "_(-1)^(0)e+bar(nu)` As X and Y both have same MASS numbers but different atomic numbers, they represent an isobar pair. |
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