The wavelength of the radiation emitted, when in hydrogen atom electron falls from infinity to stationary state $$1$$, would be (Rydberg constant $$= 1.097 \times 10^7$$ m$$^{-1}$$)

The electron is falling from $$n_2=\infty$$ to the first stationary state $$n_1=1$$ in a hydrogen atom. Such transitions belong to the Lyman series.

Using the Rydberg formula,

$$\frac{1}{\lambda}=R_HZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

For hydrogen, $$Z=1$$, $$n_1=1$$ and $$n_2=\infty$$.

Substituting these values,

$$\frac{1}{\lambda}=1.097\times10^7\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)$$

Since,

$$\frac{1}{\infty^2}=0$$

the equation becomes

$$\frac{1}{\lambda}=1.097\times10^7\ \text{m}^{-1}$$

Therefore,

$$\lambda=\frac{1}{1.097\times10^7}\ \text{m}$$

$$\lambda=0.9115\times10^{-7}\ \text{m}$$

Converting into nanometers,

$$\lambda=0.9115\times10^{-7}\times10^9\ \text{nm}$$

$$\lambda=91.15\ \text{nm}$$

Hence, the wavelength of the emitted radiation is

$$\boxed{91\ \text{nm}}$$

Thus, the correct answer is Option (A).