Answer
a) ${\bf 3.322} \;t_{1/2}$
b) ${\bf 6.644} \;t_{1/2}$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The amount of radioactive material remaining after time $ t $ is given by
$$
N = N_0 \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
$$
\frac{N}{N_0}= \left( \frac{1}{2} \right)^{t / t_{1/2}}\tag 1
$$
where:
- $ N $ is the amount of the substance remaining.
- $ N_0 $ is the initial amount of the substance.
- $ t $ is the elapsed time.
- $ t_{1/2} $ is the half-life.
If 90% has decayed, then 10% remains, so $ \frac{N}{N_0} = 0.10$.
Plug into (1);
$$
\frac{N}{N_0}= \left( \frac{1}{2} \right)^{t / t_{1/2}} =0.10
$$
Take the Natural Logarithm of Both Sides:
$$
\ln(0.10) = \frac{t}{t_{1/2}} \ln\left( \frac{1}{2} \right)
$$
Thus,
$$
t=\dfrac{ \ln(0.10) }{\ln\left( \frac{1}{2} \right)}t_{1/2}
$$
$$
t=\color{red}{\bf 3.322} \;t_{1/2}
$$
$$\color{blue}{\bf [b]}$$
If 99% has decayed, then 1% remains, so $ \frac{N}{N_0} = 0.01$.
By the same approach,
$$
\frac{N}{N_0}= \left( \frac{1}{2} \right)^{t / t_{1/2}} =0.01
$$
$$
t=\dfrac{ \ln(0.01) }{\ln\left( \frac{1}{2} \right)}t_{1/2}
$$
$$
t=\color{red}{\bf 6.644}\; t_{1/2}
$$
