JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, activity pf radioactive element at time t1,
A1=A
and at time t2,A2=A∕5
As we know that,
activity at any time(A)=A0e−λt
where A0 is activity at time t=0.
∴‌A‌=A0e−λt1 . . . (i)
‌ and ‌‌A∕5=A0e−λt2 . . . (ii)
⇒‌‌5=e−λt1+λt2
Taking log on both sides, we get
ln‌5=λ(t2−t1)‌ln‌e
⇒ ln‌5=λ(t2−t1) [∵ln‌e=1]
⇒ λ=ln‌5∕(t2−t1)
⇒ Also, average life, τ=‌

1

λ

⇒‌‌τ=‌

t2−t1

ln‌5