Calculate the de-Broglie's wavelength of an electron residing in the 2nd Bohr's orbit of a hydrogen atom. (Bohr's radius, a₀=0.529 Å )

Step-by-step solution

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. n =2 r aligned & n =2 (5.29 10⁻¹¹ ) ( n^2 Z ) & =2 (5.29 10⁻¹¹ ) n Z ...(i) aligned where, r=5.29 10⁻¹¹ ( n^2 Z ) Putting Z=1 (For H-atom) n=2 (For 2-orbit) in Eq. (i), we get aligned & =4 (5.29 10⁻¹¹ ) m & =4 0.529 10⁻¹⁰ ~m & =2.116 Å aligned Hence, de-Broglie wavelength in 2nd orbit is 2.116 Å .