Two radioactive materials `R_(1)andR_(2)` have deacy constants `6lamdaandlamda` respectively. The half life of `R_(2)" is "1.4xx10^(17)` s. Initially they contain some number of nuclei.
The time at which the ratio of the remaining nuclei of `R_(2)` to that of `R_(1)` will be e is (Let In 2=0.7)

Options

A. `2xx10^(16)s`
B. `4xx10^(16)s`
C. `3xx10^(16)s`
D. `5xx10^(16)s`

Answer

B

Explanation

For two radioactive elements ` R_(1) " and " R_(2)` decay contants,
` lambda_(1) = 6lambda " and " lambda_(2) = lambda `
Half-life of ` R_(2) (t_(1//2)_(2) = 1.4xx 10^(17)s `
If ` N_(0)` be the initial number of nuclei at t = 0 , in both
` R_(1)" and " R_(2) " or " N_(1) " and " N_(2)` be the nuclei after time t.
Then, ` N_(1) = N_(0) ((1)/(2))^(n1)`
and ` N_(2) = N_(0) ((1)/(2))^(n2)`
` therefore (N_(2))/(N_(1)) = ((1)/(2))^(n2-n1)`
` e = ((1)/(2))^(n2 - n1)" " [ because (N_(2))/(N_(1)) = e]`
` log_(e) e = (n_(2) - n_(1)) " log" (1)/(2)`
` 1 = (n_(2) - n_(1)) " log"2^(-1)`
` rArr 1 = (n_(1) - n_(2)) " log" 2 `
` 1 = (n_(1) - n_(2)) 0.7`
` (t)/((t_(1//2))_(1)) - (t)/((t_(1//2))_(2)) = (1)/(0.7)`
Since , ` (t_(1//2))_(2) = (0.693)/(lambda_(2)) = (0.693)/(lambda)`
` rArr lambda = (0.693)/((t_(1//2))_(2)) = (0.693)/(1.4xx10^(17)) = 0.495 xx 10^(-17)`
` therefore lambda_(1) = 6 lambda = 6 xx 0.495 xx 10^(-17) `
` = 2.97 xx 10^(-17) m`
` therefore (t_(1//2))_(1) = (0.693)/(lambda_(1)) = (0.693)/(2.97 xx 10^(-17))`
` (t_(1//2))_(1) = 0.23 xx 10^(17) s `
` therefore ` From Eq. (i) , we get
` (t) /(0.23 xx 10^(17)) - (t)/(1.4 xx 10^(17)) = (1)/(0.7)`
` t ((1)/(0.23) - (1)/(1.4)) = (10^(17))/(0.7) `
` t (3.64 ) = (10^(17))/(0.7) `
` t = (10^(17))/(0.7 xx 3.64 ) = (10)/(2.56) xx 10^(16)`
` = 3.9 xx 10^(16) ~~ 4 xx 10^(16) s `