If the radius of first orbit of H atom is $$a_0$$, the deBroglie wavelength of an electron in the third orbit is

For an electron in the $$n^{\text{th}}$$ Bohr orbit, the de Broglie wavelength $$\lambda$$ satisfies the condition that an integral number of wavelengths fits exactly on the circumference:

$$2\pi r_n = n\,\lambda \quad -(1)$$

The radius of the $$n^{\text{th}}$$ orbit for a hydrogen atom is given by Bohr’s formula:

$$r_n = n^{2} a_0 \quad -(2)$$

Substitute $$r_n$$ from $$-(2)$$ into the quantization condition $$-(1)$$ and solve for $$\lambda$$:

$$2\pi (n^{2} a_0) = n\,\lambda$$

$$\lambda = \frac{2\pi n^{2} a_0}{n} = 2\pi n a_0 \quad -(3)$$

For the third orbit, $$n = 3$$. Using $$-(3)$$:

$$\lambda_3 = 2\pi \times 3 \times a_0 = 6\pi a_0$$

Thus, the de Broglie wavelength of an electron in the third orbit is $$6\pi a_0$$.

Option C which is: $$6\pi a_0$$