Question

A radioactive material of half-life time of 69.3 days kept in a container $\frac{2}{3}rd$ of the substance remains undecayed after (given, ln $\frac{3}{2}=0.4$)

• A. 20 days
• B. 25 days
• C. 35 days
• D. 40 days
• E. 50 days

Answer 1

The correct option is D 40 days

Here, decay constant $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{69.3}=0.01day{s}^{-1}$

By radio activity law, $N=N_0{e}^{-\lambda t}$

or $e^{\lambda t}=\frac{N_0}{N}$

Taking $ln$, $\lambda t=ln\left(N_0/N\right)=ln\left(\frac{N_0}{\left(2/3\right)N_0}\right)=ln\left(3/2\right)=0.4$

or $t=0.4/\lambda =0.4/0.01=40$ days