The rate of decay of a radioactive species is given by `R_(1)` at time `t_(1)` and `R_(2)` at later time `t_(2)`. The mean life of this radioactive species is:

Options

A. `(t_(2) - t_(1))/("In" R_(1) - "In" R_(2))`
B. `(t_(2) - t_(1))/("In" R_(2) - "In" R_(1))`
C. `(t_(2) + t_(1))/("In" R_(2)+ "In" R_(1))`
D. `(t_(2) - t_(1))/("In" R_(2)+ "In" R_(1))`

Answer

`(a)`

Explanation

`R_(1) = lambdaN_(1), R_(2) = lambda N_(2)`
`:. (R_(1))/(R_(2)) = (N_(1))/(N_(2))`
Also `t_(1) = (1)/(lambda) log_(e) ((N_(0))/(N_(1)))`
`t_(2) = (1)/(lambda) log_(e) ((N_(0))/(N_(2)))`
`:. T_(1) - t_(2) = (1)/(lambda) [log_(e) ((N_(0))/(N_(1))) - log_(e) ((N_(0))/(N_(2)))]`
`= (1)/(lambda) log_(e) ((N_(2))/(N_(1)))`
`= (1)/(lambda) log_(e) ((R_(2))/(R_(1)))`
`T = (1)/(lambda) = (t_(1) - t_(2))/(log_(e) R_(2) - log_(e) R_(1))`