Discussion Forum

 The half-life of radioactive Radon is 3.8 days. the time at the end of which 1/20 th of the radon sample will remain undecayed is (given  $\log_{10}e=0.43434 $)

Answer:

Explanation:

t_1/2  =3.8 day

 $\therefore$    $\lambda= \frac{0.693}{t_{1/2}}=\frac{0.693}{3.8}=0.182$

 If the initial number of atom is a = A₀ then after time r the number of atoms is a/20= A , we have to find t.

 $t= \frac{2.303}{\lambda}\log\frac{A_{0}}{A}=\frac{2.303}{0.182}\log\frac{a}{a/20}$

  $=\frac{2.303}{0.182}\log20=16.46$  days