Question

A sample of a radioactive substance undergoes 80% decomposition in 345 minute. its half life is?

Answer 1

K= 2.303/t x log a/ a-x t = 345mnts a= 100(say) x = 80 Therefore k= 2.303/345 log 100 / 100-80 K= 2.303/345 log 100/20= 0.00466 Hence half life= 0.693/k = 148.7 minutes

Answer 2

A sample of a radioactive substance undergoes 80% decomposition in 345 minutes.

Let initially, $N_0$ amount of radioactive substance is taken.

after t = 345 minutes, amount of substance , $N_t=N_0-80\%\:\textbf{of}\:N_0=0.2N_0$

using radioactive decay formula,

$N_t=N_0e^{-\lambda t}$

or, $0.2N_0=N_0e^{-\lambda 345}$

or, $ln(0.2) =-345\lambda$

or, $\lambda=\frac{-ln(0.2)}{345}=\frac{ln5}{345}$ ......(1)

[ -ln(0.2) = ln(0.2)^-1 = ln(1/0.2) = ln5]

now,half life , $T_{1/2}=\frac{ln2}{\lambda}$

from equation (1),

$T_{1/2}=\frac{345ln2}{ln5}$

putting ln2 = 0.693 and ln5 = 1.609

so, half life , $T_{1/2}$=345 × 0.693/1.609 = 148.6 min