The de-Broglie wavelength ($$\lambda_B$$) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($$\lambda_G$$) by:

For an electron moving in a Bohr orbit we begin with the de-Broglie standing-wave condition. The condition states that an integral number of de-Broglie wavelengths must exactly fit along the circumference of the circular orbit. Mathematically, the statement is written as

$$2\pi r_n \;=\; n\,\lambda_n,$$

where $$r_n$$ is the radius of the $$n^{\text{th}}$$ Bohr orbit, $$\lambda_n$$ is the de-Broglie wavelength of the electron in that orbit, and $$n$$ is the principal quantum number.

Rearranging this relation to make $$\lambda_n$$ the subject we obtain

$$\lambda_n \;=\; \frac{2\pi r_n}{n}.$$

Now we substitute the Bohr‐model expression for the orbit radius. In the hydrogen atom, the radius of the $$n^{\text{th}}$$ orbit is

$$r_n \;=\; n^2 a_0,$$

where $$a_0$$ is the Bohr radius, a constant equal to $$0.529\;\text{Å}$$. Substituting this value of $$r_n$$ into the previous formula gives

$$\lambda_n = \frac{2\pi (n^2 a_0)}{n}.$$

Carrying out the division by $$n$$ in the numerator, we find

$$\lambda_n = 2\pi a_0\,n.$$

This result is very revealing: the de-Broglie wavelength of the electron in a hydrogenic Bohr orbit is directly proportional to the principal quantum number $$n$$. Symbolically,

$$\lambda_n \propto n.$$

We are asked to compare the wavelength in the second excited state with that in the ground state. The ground state corresponds to $$n = 1$$, and the second excited state corresponds to $$n = 3$$ (because the first excited state is $$n = 2$$ and the second excited state is the next one up, $$n = 3$$).

Let $$\lambda_G$$ denote the wavelength in the ground state and $$\lambda_B$$ denote the wavelength in the second excited state. Using the proportionality just obtained, we write

$$\frac{\lambda_B}{\lambda_G} = \frac{n_B}{n_G}.$$

Substituting $$n_B = 3$$ and $$n_G = 1$$, we get

$$\frac{\lambda_B}{\lambda_G} = \frac{3}{1} = 3.$$

Multiplying both sides by $$\lambda_G$$ finally yields

$$\lambda_B = 3\,\lambda_G.$$

Hence, the correct answer is Option A.