The half-life of a radioactive substance is `30` minutes, The time (in minutes) taken between `40 %` decay and `85 %` decay of the same radioactive substance is.

Options

A. 45
B. 60
C. 15
D. 30

Explanation

Using equation of radioactivity
`N_(1)=N_(2)e^(-lamdat)`
40% decay
`60N_(2)e^(-lamdat)`………1
85% decay
`15=N_(0)e^(-lamdat_(s))` …..2
Dividing eq. 1 by eq. 2 we get
`60/15=(e^(-lamdat_(1))/(e^(-lamdat_(2)))implies4=e^(????)`
Take log both side
`log4=loge^(???)`
`2In2=lamda(t_(2)-t_(1))" ":'lamda=(In2)/(t_(1//2))`
`2In2=(In2)/30(t_(2)-t_(1))`
`t_(2)-t_(1)=60` minutes