In a hydrogen atom, what is the de-Broglie wavelength of an electron in the second Bohr orbit? [Given that Bohr radius $a_0=52.9\text{pm}$ ]

According to Bohr's quantization condition, the circumference of the $n$ -th orbit is an integral multiple of the de-Broglie wavelength:$2\pi r_n=n\lambda$ For the second orbit ($n=2$):$2\pi r_2=2\lambda ⟹\lambda =\pi r_2$ The radius of the $n$ -th orbit is $r_n=a_0{n}^2$.For $n=2$, $r_2=52.9\times 2^2=52.9\times 4=211.6\text{pm}$.Therefore, $\lambda =211.6\pi \text{pm}$.