1. In Bohr's model, the electron in the $n^{\text{th}}$ orbit satisfies the standing wave condition:

$2\pi r_n = n\lambda_n$

where $\lambda_n$ is the de-Broglie wavelength of the electron in that orbit.

2. Therefore,

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

Also, for hydrogen atom in Bohr model,

$r_n \propto n^2$

Hence,

$\lambda_n \propto \frac{n^2}{n} = n$

So the de-Broglie wavelength is directly proportional to the principal quantum number $n$:

$\lambda_n \propto n$

3. Ground state corresponds to $n=1$.

So,

$\lambda_G = \lambda_1$

4. Second excited state means the electron is in the third orbit:

• Ground state $\to n=1$
• First excited state $\to n=2$
• Second excited state $\to n=3$

Thus,

$\lambda_B = \lambda_3$

Since $\lambda_n \propto n$,

$\frac{\lambda_3}{\lambda_1} = \frac{3}{1}$

Therefore,

$\lambda_B = 3\lambda_G$

5. Checking options:

• A: $\lambda_B = 2\lambda_G$ $\rightarrow$ Incorrect
• B: $\lambda_B = 3\lambda_G$ $\rightarrow$ Correct
• C: $\lambda_B = \lambda_G/2$ $\rightarrow$ Incorrect
• D: $\lambda_B = \lambda_G/3$ $\rightarrow$ Incorrect

Hence, the correct option is B.