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In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: [Given that Bohr radius, $a_0=52.9$ pm]
Detailed Solution
$n\lambda=2\pi r$ with $r=\frac{n^2}{Z}a_0$. For $n=2$: $2\lambda=2\pi\times4\times52.9 \Rightarrow \lambda=2\pi\times2\times52.9=211.6\pi$ pm.