The de-Broglie wavelength λ B associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state λ G by:

Step-by-step solution

We know that wavelength is inversely proportional to momentum, let λ B is de-Broglie Wavelength corresponds to momentum P B , similarly let λ G is Wavelength at ground state corresponds to momentum P G , λ B λ G =   P G P B =   m v G m v B , and velocity, V ∝ z n , where Z is the atomic number, and n is a number of orbits, here, n G = 1 (ground state), n B = 3 (second excited state). So, λ B λ G =   n B n G =   3 1 ⇒ λ B = 3 λ G