A solution contains a mixture of isotopes of X""^(A""_(1))(t""_(1//2)=14 days) and X""^(A""_(2))(t""_(1//2)=25days). Total activity is 1 curie at t=0. The activity reduces by 50% in 20 days. Find : a) The initial activities of X""^(A""_(1)) and X""^A""_(2)). b) The ratio of their initial number of nuclei.

A solution contains a mixture of isotopes of X""^(A""_(1))(t""_(1//2)=

1.	A solution contains a mixture of isotopes of X""^(A""_(1))(t""_(1//2)=14 days) and X""^(A""_(2))(t""_(1//2)=25days). Total activity is 1 curie at t=0. The activity reduces by 50% in 20 days. Find : a) The initial activities of X""^(A""_(1)) and X""^A""_(2)). b) The ratio of their initial number of nuclei.
Answer» Solution :Let activity of `X""_(A""_(1))ANDX""_(A""_(2))` are a and B curie respectively at t = 0 `therefore` a + b =1 curie …(1) Now rate `mu` Number of atoms `therefore`For `X""^(A""_(1))" "t=(2.303)/(k)log""_(10)"(N""_(0))/(N)=(2.303)/(k)log""_(10).(r""_(0))/(r)` `20=(2.303xx14)/(0693)log""_(10).(a)/(r""_(1))` `thereforer""_(1)=0.3716a` For `X""^(A""_(2))""t=(2.303)/(k)log""_(10).(N""_(0))/(N)=(2.303)/(k)log""_(10).(r""_(0))/(r)` `20=(2.303xx25)/(0.693)log""_(10).(b)/(r""_(2))` , `r""_(2)=0.5744b` GIVEN activity after 20 days `=(1)/(2)`cruie 03716a+ 0.5744b `=(1)/(2)` or `0.7432a + 1.1488b = 1""...(2)` Byequation (1) and (2) a = 0.3669 Ci = 0.3669 `xx 3.7xx10""^(10)` dps b = 0.6331 Ci = 0.6331 `xx3.7xx10""^(10)` dps Now rate = k.N`because` a = 03669 curie For 0.3669 `xx10""^(10)xx` 3.7 = `(0.693)/(14xx24xx60xx60)N""_(0)""^(A""_(1))` For 0.6331 `xx10""^(10)xx` 3.7 =`(0.693)/(25xx24xx60xx60)N""_(0)""^(A""_(2))` `therefore(N""_(0)""^(A""_(1)))/(N""_(0)""^(A""_(2)))=0.3245`