Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An exmple is: `""^(38)"S" overset ("half-life") underset(=2.48) to ""^(38)Cl overset ("half-life") underset(=0.62h) to ""^(38)Ar ("stable")` Assume that we start with 1000 `""^(38)S` nuclei at time t=0. The number of `""^(30)Cl` is of count zero at t=0 and will again be zero at `t=oo`. At what value of t, would the number of counts be a maximum?

Sometimes a radioactive nucleus decays into a nucleus which itself is

1.	Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An exmple is: `""^(38)"S" overset ("half-life") underset(=2.48) to ""^(38)Cl overset ("half-life") underset(=0.62h) to ""^(38)Ar ("stable")` Assume that we start with 1000 `""^(38)S` nuclei at time t=0. The number of `""^(30)Cl` is of count zero at t=0 and will again be zero at `t=oo`. At what value of t, would the number of counts be a maximum?
Answer» The given decay sequence is `.^(38)("Sulper") overset ("half-life") underset(=2.48) to .^(38)Cl overset ("half-life") underset(=0.62h) to .(38)Ar ("stable")` At any time t, suppose `.^(38)S` has `N_(1) (t)` active nuclei and `.^(38)Cl` have `N_(2)(t)` active nuclei. `:. (dN_(1))/(dt)=-lambda_(1)N_(1)`=rate of formation of `.^(38)Cl` and `(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)`= net rate of decay of `.^(38)Cl=-lambda_(2)N_(2)+lambda_(1)N_(0)e^(-lambda_(1)t)` Multiplying both sides by `e^(lambda_(2)t)dt` and rearranging, `e^(lambda_(2)t)dN_(2)+lambda_(2)N_(2)e^(lambda_(2)t)d t=lambda_(1)N_(0)e^((lambda_(2)-lambda_(1)t)dt` Integrating both sides, we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^((lambda_(2)-lambda_(1)t)+C.......(i)` Where C is constant of integration. At `t=0, N_(2)=0 :. C=-(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))` Putting in (i), we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^((lambda_(2)-lambda_(1)t))-1)]` `N_(2)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^(-lambda_(1))-e^(-lambda_(2)t)]` for maximum count , `N_(2)=max, (dN_(2))/(dt)=0` `(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(1)t) (-lambda_(1))=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(2)t) (-lambda_(2))` or `(lambda_(1))/(lambda_(2))=(e^(-lambda_(2)t))/(e^(-lambda_(2)t))=e^((lambda_(1)-lambda_(2))t)` or `log_(e) ((lambda_(1))/(lambda_(2)))=(lambda_(1)-lambda_(2))tlog_(e) e` or `t=(log_(e)(lambda_(1)//lambda_(2)))/((lambda_(1)-lambda_(2)))=(log_(e) T_(2)//T_(1))/(0.693(1/(T_(1))-1/(T_(2))))` `t=(2.303log_(10) (0.62//2.48)xxT_(1)T_(2))/(0.693 (T_(2)-T_(1)))=(2.303(0-0.602)xx2.48xx0.62)/(0.693(0.62-2.48))=(2.303xx0.602xx2.48xx0.62)/(0.693xx1.86)=1.65sec`

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Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An exmple is: `""^(38)"S" overset ("half-life") underset(=2.48) to ""^(38)Cl overset ("half-life") underset(=0.62h) to ""^(38)Ar ("stable")` Assume that we start with 1000 `""^(38)S` nuclei at time t=0. The number of `""^(30)Cl` is of count zero at t=0 and will again be zero at `t=oo`. At what value of t, would the number of counts be a maximum?

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