Discussion Forum

A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10³  disintegrations per second. Given that ln 2=0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80s after preparation of the sample is 

Answer:

Explanation:

Number of nuclei decayed in time t,

 $N_{d}=N_{0}(1-e^{-\lambda t)}$

  $\therefore$    % decayed=   $=\left(\frac{N_{d}}{N_{0}}\right)\times 100$

    =    $(1-e^{-\lambda t}d)$ x100     ......(i)

  Here,   $\lambda=\frac{0.693}{1386}=5 \times10^{-4}s^{-1}$

  $\therefore$   % decayed=   $(\lambda t)\times100$

  =  $(5 \times 10^{-4})(80)(100)=4$