## In This post you will get answer of Q.

A radioactive isotope has a half-life of 2 yr. How long will it take the activity to reduce to 3% of its original value?

Solution:

## Number of atoms remained undecayed after time t, $ N = N_0 e^{- lambda t} $

Given, $ N = frac{3N_0}{100} $

$ therefore frac{3N_0}{100} = N_0 e^{- lambda t}, , or, , 3 = 100 e^{- lambda t}, , , ……..(i) $

Now, $ lambda = frac{0.693}{T_{1/2}} = frac{0.693}{2} = 0.346 $

Substituting the value of $ lambda $ in Eq, (i), we get

$ 2.303 log bigg(frac{3}{100}bigg) = – 0.346 t $

$ – 1.52 times 2.303 = – 0.346 t $

$ Rightarrow t = 10 yr $