The de-Broglie wavelength (λB) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λG) by:
λB=λG/3
λB=2λG
λB=3λG
λB=λG/2
Solution:
The correct option is D λB=3λG de- Broglie wavelength λ=h/mv By conservation of angular momentum, mvr=nh/2π h/mv=2πrn/n λ=2πr r=a₀n²/Z λ=2πa₀n/Z As the atom is hydrogen Z=1 λB=2πa₀(3)/1 λG=2πa₀(1)/1 λB=3λG