In a hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: [Given that Bohr radius $a_0=52.9$ pm]

Solution & Explanation

\textbf{Answer:} (B) $211.6\pi$ pm \textbf{Solution:} Bohr quantization gives $n\lambda =2\pi r_n$. Here $r_n=a_0\frac{n^2}{Z}=52.9\times 4=211.6$ pm. So $\lambda =\frac{2\pi r_n}{n}=\frac{2\pi \times 211.6}{2}=211.6\pi$ pm.