The radioactivity of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at a time, `T_(2)` If the half-line of the secimen is `T`, the number of atoms that have disintergrated in the time `(T_(2) - T_(1))` is proportional to:

Options

A. `(R_(1) T_(1) - R_(2)T_(2))`
B. `(R_(2) - R_(1))`
C. `(R_(2) - R_(1))//T`
D. `(R_(2) - R_(1))T//0.693`

Answer

`(d)`

Explanation

`R_(1) = lambda N_(1)`,
`R_() = lambda N_(2)`.
No, of atoms decayed in time
`(T_(2) - T_(1)) = N_(1) - N_(2)`
`= (R_(1) - R_(2))/(lambda) = (R_(1) - R_(2))/(0.693) T.`