According to the de Broglie relation and Bohr's postulate of angular momentum quantization, the circumference of the $n$-th orbit is an integral multiple of the de Broglie wavelength ($\lambda$). The relationship is given by: $2\pi r_n = n\lambda$ where $r_n$ is the radius of the $n$-th orbit. The radius of the $n$-th orbit in a hydrogen atom is given by the formula: $r_n = n^2 a_0$ For the second Bohr orbit, $n = 2$. Given $a_0 = 52.9$ pm: $r_2 = (2)^2 \times 52.9 \text{ pm} = 4 \times 52.9 \text{ pm} = 211.6 \text{ pm}$ Substituting the values into the circumference equation: $2\pi (211.6) = 2 \lambda$ $\lambda = \pi \times 211.6 \text{ pm} = 211.6 \pi \text{ pm}$