At a given, t = 0, two radioactive substances A and B have equal activities. The ratio `(R_(B))/(R_(A))` of their activities after time t itself decays with time t as `e^(-3t)`, If the half-life of A is ln2, the half-life of B is :

Options

A. `("In 2")/4 `
B. 2 In 2
C. 4 In 2
D. `("In 2")/2 `

Answer

A

Explanation

`A = lambdaN `
Initially `N = N_(0)`
`A_(A) = A_(B)`
`lambda_(A)N_(0_(A)) =lambda_(B)N_(0_(B))`
`(T_(1//2))_(A) = ("In(2)")/(lambda_(A)) rArr In(2) = ("In(2)")/(lambda_(A))`
`rArr lambda_(A) = 1 `
At time t ,
`(A_(B))/(A_(A)) =e^(-3t) =(lambda_(B)N_(0_(B))e^(-lambda_(B)t))/(lambda_(A)N_(0_(A))e^(-lambda_(A)t))rArr e^(-3t) = e^((lambda_(A)-lambda_(B))t)`
` -3 = lambda_(A) -lambda_(B) rArr lambda_(B) = 1+3 =4`
`(T_(1//2))_(B) = ("In(2)")/(lambda_(B))= ("In(2)")/4`