Consider the first order reaction $$R \to P$$. The fraction of molecules decomposed in the given first order reaction can be expressed as :

For a first-order reaction $$R \rightarrow P$$, the rate law is written as

$$\text{Rate} = -\frac{d[R]}{dt} = k_1 [R]$$

Rearrange and integrate between the limits $$[R]_0$$ at $$t = 0$$ and $$[R]_t$$ at any time $$t$$:

$$\int_{[R]_0}^{[R]_t} \frac{d[R]}{[R]} = -k_1 \int_{0}^{t} dt$$

This gives

$$\ln\!\left(\frac{[R]_t}{[R]_0}\right) = -k_1 t$$

Exponentiate both sides:

$$\frac{[R]_t}{[R]_0} = e^{-k_1 t}$$

Hence the concentration of reactant remaining after time $$t$$ is

$$[R]_t = [R]_0\, e^{-k_1 t}$$

The fraction of molecules decomposed (i.e., that have reacted to form $$P$$) is

$$\text{Fraction decomposed} = \frac{[R]_0 - [R]_t}{[R]_0}$$

Substitute $$[R]_t$$ from above:

$$\text{Fraction decomposed} = \frac{[R]_0 - [R]_0 e^{-k_1 t}}{[R]_0} = 1 - e^{-k_1 t}$$

Therefore, the correct expression is $$1 - e^{-k_1 t}$$.

Option D which is: $$1 - e^{-k_1 t}$$