How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to 1/8^th of its initial value?

 we know that , N=N₀e^{-lambda * t}

and t_1/2=ln2/lambda , so lambda=ln2/t_1/2 , lamda = ln2/T ______(1)

now , N=N0/8 (Given .)

N0/8=N₀ e^{-ln2/T * t} 

1/8 = e^{-ln2/T * t} 

e^{ln2/T * t} = 8

Taking log both sides , ln8=ln2/T * t

  3ln2 = ln2/T * t

  T = 3T