Bohr's radius is equal to most probable distance between the nuclear and electron in H-atom in its ground states. According to de-Broglie wavelength, the allowed any stationary orbit i.e. nλ=2πr nλ=2π(5.29×10−11)(
n2
Z
) λ=2π(5.29×10−11)×
n
Z
...(i) where, r=5.29×10−11(
n2
Z
) Putting Z=1 (For H-atom) n=2 (For 2-orbit) in Eq. (i), we get λ=4π(5.29×10−11)m λ=4×0.529π×10−10m λ=2.116πÅ Hence, de-Broglie wavelength in 2nd orbit is 2.116πÅ.