Solve the E question on half life

The half-life ($t_{1/2}$) of the radioactive isotope is 15 days. We need to find the time ($t$) it takes for 90% of the isotope to decay. This means 10% of the isotope remains.

If 90% of the isotope has decayed, then the fraction remaining is $100\% - 90\% = 10\%$, or 0.10.
So, $\frac{N_t}{N_0} = 0.10$, where $N_t$ is the amount remaining at time $t$, and $N_0$ is the initial amount.

The formula relating the amount remaining to the initial amount and time is:

Divide by $N_0$:

Take the natural logarithm (ln) of both sides:

Using the logarithm property $\ln(a^b) = b \ln(a)$:

We know that $\ln\left(\frac{1}{2}\right) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2)$.
So,

Rounding to a reasonable number of significant figures:

The time it will take for 90% of the isotope to decay is approximately $\boxed{\text{49.8 days}}$.