# In sample of hydrogen atom in ground state electron make transition from ground state to particular excited state where path length is 5times de broglie wavelength, electron make back transition to ground state producing all possible photons. If photon having 2nd highest energy of sample can be used to excite the electron in a particular excited state of Li2+ atom then find final excited state of Li2+ atom

509 Points
12 years ago

let this state is n then

de broglie wavelength = h/mv

length of path = 2pir                 ( r is radius)

length of path = 5 (de broglie wavelength)

2pir = 5h/mv

mvr = 5h/2pi  ..................1

we have  , mvr = nh/2pi        .............2                   (bohrs quantixation of angular momentam)

comparing both

n = 5

now , maximum energy = 13.6(1-1/25)= 13.056

second maximum energy = 13.6(1-1/16)           ...................3

for Li2+ , E = 13.6(1/n12 - 1/n22)*9                           (Z  = 3 )

this energy is equal to energy from in 3

13.6(1-1/16) = 13.6*9(1/n12 - 1/n22)

1/9 - 1/9*16 = 1/32 - 1/122 =  1/n12 - 1/n22

n1 = 3 , n2 = 12

final exited state = n = 12