Bohr radius, $a_{0}=52.9 pm$ $n=2, r_{n}=n^{2} a_{0}=(2)^{2} a_{0}=4 \times 52.9 pm =211.6 pm$ The angular momentum of an electron in a given stationary state can be expressed as in equation, $m v r=n \cdot \frac{h}{2 \pi}=2 \times \frac{h}{2 \pi}=\frac{h}{\pi}$
de-Broglie equation mvr $\pi=h$ $\lambda=\frac{h}{m v} ; \hat{\lambda} m v=h$
From equatisns, (i) and (ii), we get $\lambda=\pi r$
Putting the value of $r, \lambda=211.6 \pi pm$