Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $$\lambda$$. The value of principal quantum number $$n$$ of the excited state will be: ($$R$$ : Rydberg constant)

When a hydrogen atom transitions from an excited state with principal quantum number $$n$$ to the ground state ($$n_f = 1$$), the wavelength of the emitted photon is given by the Rydberg formula:

$$\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{n^2}\right)$$

where $$R$$ is the Rydberg constant and $$\lambda$$ is the wavelength of the emitted photon.

Expanding:

$$\dfrac{1}{\lambda} = R - \dfrac{R}{n^2}$$

Isolating the term with $$n$$:

$$\dfrac{R}{n^2} = R - \dfrac{1}{\lambda}$$

Taking $$\lambda$$ as the common denominator on the right side:

$$\dfrac{R}{n^2} = \dfrac{\lambda R - 1}{\lambda}$$

Solving for $$n^2$$:

$$n^2 = \dfrac{R \cdot \lambda}{\lambda R - 1}$$

Taking the square root:

$$n = \sqrt{\dfrac{\lambda R}{\lambda R - 1}}$$

Therefore, the correct answer is Option B.