Carbon ( Z = 6 ) possessing mass number 11 decays into boron ( Z =...

Let the total number of atoms in the mixture be $N$.

Initially, the number of Carbon-11 atoms is $N_1 = 0.90 N$.

Finally, the number of Carbon-11 atoms is $N_2 = 0.10 N$.

According to the law of radioactive decay:

$N_2 = N_1 e^{-\lambda t}$

$0.10 N = 0.90 N e^{-\lambda t}$

$e^{\lambda t} = 9$

Taking the natural logarithm on both sides:

$\lambda t = \ln 9 = 2 \ln 3$

Since the decay constant $\lambda = \dfrac{\ln 2}{T_{1/2}}$, we get:

$t = \dfrac{2 \ln 3}{\ln 2} T_{1/2}$

Substituting $T_{1/2} = 20.3 \text{ minutes}$:

$t = \dfrac{2 \times 0.4771}{0.3010} \times 20.3$

$t \approx 3.17 \times 20.3 \approx 64.35 \text{ minutes}$

The closest value is $64.3 \text{ minutes}$.