The de-Broglie wavelength associated with the electron in the $$n = 4$$ level is:

We know from Bohr’s model that an electron in the $$n^{\text{th}}$$ stationary orbit of the hydrogen atom moves in such a way that its de-Broglie wavelength $$\lambda_n$$ fits an integral number of times around the circumference of that orbit. Mathematically the de-Broglie condition is stated as

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

Here $$r_n$$ is the radius of the $$n^{\text{th}}$$ orbit and $$n$$ is the principal quantum number.

Bohr also gave the expression for the radius of the $$n^{\text{th}}$$ orbit:

$$r_n = n^2 a_0$$

where $$a_0$$ is the Bohr radius (a constant equal to about $$0.53\ {\rm Å}$$).

Now we substitute the value of $$r_n$$ from the second formula into the first one:

$$2\pi \Bigl(n^2 a_0\Bigr) = n\lambda_n$$

We simplify step by step:

$$2\pi n^2 a_0 = n\lambda_n$$

Divide both sides by $$n$$ to isolate $$\lambda_n$$:

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

which reduces to

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

We can now see very clearly that the de-Broglie wavelength is directly proportional to the principal quantum number $$n$$:

$$\lambda_n \propto n$$

Let us write the explicit values for the ground state $$n = 1$$ and for the excited state $$n = 4$$.

For the ground state:

$$\lambda_1 = 2\pi (1) a_0 = 2\pi a_0$$

For the fourth orbit:

$$\lambda_4 = 2\pi (4) a_0 = 8\pi a_0$$

Next we compare $$\lambda_4$$ with $$\lambda_1$$:

$$\frac{\lambda_4}{\lambda_1} = \frac{8\pi a_0}{2\pi a_0} = 4$$

This gives

$$\lambda_4 = 4\lambda_1$$

In words, the de-Broglie wavelength of the electron in the $$n = 4$$ level is four times the de-Broglie wavelength of the electron in the ground state.

That matches Option B in the given list.

Hence, the correct answer is Option B.