If the radius of the first orbit of hydrogen atom is $$a_0$$, then de Broglie's wavelength of electron in 3$$^{rd}$$ orbit is

The radius of the $$n$$th orbit of hydrogen atom is:

$$ r_n = n^2 a_0 $$

For the 3rd orbit: $$r_3 = 9a_0$$.

By Bohr's quantization condition, the circumference of the orbit equals $$n$$ times the de Broglie wavelength:

$$ 2\pi r_n = n\lambda $$

For $$n = 3$$:

$$ 2\pi(9a_0) = 3\lambda $$

$$ \lambda = \frac{18\pi a_0}{3} = 6\pi a_0 $$

The de Broglie wavelength of the electron in the 3rd orbit is $$6\pi a_0$$.