The half-life period of a first-order reaction is 15 minutes. The amount of substance left after one hour will be:

The half-life period of a first-order reaction is given as 15 minutes. For a first-order reaction, the half-life ($$ t_{1/2} $$) is related to the rate constant ($$ k $$) by the formula:

$$ t_{1/2} = \frac{\ln 2}{k} $$

Substituting the given half-life:

$$ 15 = \frac{\ln 2}{k} $$

Solving for $$ k $$:

$$ k = \frac{\ln 2}{15} $$

The amount of substance left after time $$ t $$ for a first-order reaction is given by:

$$ N = N_0 e^{-kt} $$

where $$ N_0 $$ is the original amount and $$ N $$ is the amount left at time $$ t $$. We need to find the amount after one hour, which is 60 minutes. So, $$ t = 60 $$ minutes.

Substitute $$ k $$ and $$ t $$:

$$ N = N_0 e^{-\left( \frac{\ln 2}{15} \right) \times 60} $$

Simplify the exponent:

$$ \frac{\ln 2}{15} \times 60 = \ln 2 \times \frac{60}{15} = \ln 2 \times 4 = 4 \ln 2 $$

So the equation becomes:

$$ N = N_0 e^{-4 \ln 2} $$

Using the property $$ e^{c \ln a} = a^c $$, we get:

$$ e^{-4 \ln 2} = (e^{\ln 2})^{-4} = 2^{-4} $$

Therefore:

$$ N = N_0 \times 2^{-4} = N_0 \times \frac{1}{2^4} = N_0 \times \frac{1}{16} $$

Alternatively, using the half-life concept: the number of half-lives in 60 minutes is $$ \frac{60}{15} = 4 $$. After each half-life, the amount halves, so after 4 half-lives, the amount is $$ \left( \frac{1}{2} \right)^4 = \frac{1}{16} $$ of the original amount.

Hence, the amount left after one hour is $$ \frac{1}{16} $$ of the original amount. Comparing with the options, Option A corresponds to $$ \frac{1}{16} $$ of the original amount.

Hence, the correct answer is Option A.