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

Options

A. `lamda_B = lamda_(g//3)`
B. `lamda_B = lamda_(G//2)`
C. `lamda_(B ) = 2lamda_(G)`
D. `lamda_(B ) = 3 lamda_(G)`

Answer

D

Explanation

De -Brolie wavelength `,lamda = (h)/(p)`
`(lamda_B )/(lamda_G)=(P_a)/(p_B)=(mv _G )/(mv_B)`
speed of electron ` V prop (z )/(n ) `
so ` (lamda_B )/(lamda_G)=(n_B )/( n_G) = (3)/(1 ) implies lamda_b = 3 lamda _G`