A hydrogen like atom (atomic number Z) is uin a higher excleted atate of quantum n , The excited atom can make a two photon of energy `10.2 and 17.0 eV` respactively , Alernately the atom from the same excited state by successively eniting two photons of energies `4.25 eV and 5.95 eV` respectively
Determine the value of n and Z (lonization energy of H- atom `= 13.6 eV`)
A hydrogen like atom (atomic number Z) is uin a higher excleted atate of quantum n , The excited atom can make a two photon of energy `10.2 and 17.0 eV` respactively , Alernately the atom from the same excited state by successively eniting two photons of energies `4.25 eV and 5.95 eV` respectively
Determine the value of n and Z (lonization energy of H- atom `= 13.6 eV`)
Determine the value of n and Z (lonization energy of H- atom `= 13.6 eV`)
Text Solution
Verified by Experts
The correct Answer is:
C
For hydrogen like atoms
E_(0) = (13.6)/(n^(2)) Z^(2) eV//atom`
`Given E_(n) - E_(2) = 10.2 + 17 = 27.2 eV` ….(i)
`E_(n) - E_(3) = 4.24 + 5.95 = 10.2 eV`
`:. E_(3) - E_(2) = 17`
`But E_(3) - E_(2) = (13.6)/(9)Z^(2) - ((-13.6)/(4)Z^(2))`
`= - 13.6 Z^(2) [(1)/(9) - (1)/(4)]`
`= - 13.6 Z^(2) [(4 - 9)/(36)] = (13.6 xx 5)/(36)Z^(2)`
` (13.6 xx 5)/(36)Z^(2) = 17 rArr Z= 3
`E_(n) - E_(2) = - (13.6)/(n^(2)) xx 3^(2) - [-(13.6)/(2^(2) xx 3^(2) ]`
`- 13.6 [(9)/(n^(2)) - (9)/(4)] = -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]` ....(ii)
frequency (i) and (ii)
` -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]= 27.2`
`rArr 122.4 (4 - n^(2)) = 108.8 n^(2)`
`rArr n^(2)= (489)/(13.6) = 36 rArr n = 6`
E_(0) = (13.6)/(n^(2)) Z^(2) eV//atom`
`Given E_(n) - E_(2) = 10.2 + 17 = 27.2 eV` ….(i)
`E_(n) - E_(3) = 4.24 + 5.95 = 10.2 eV`
`:. E_(3) - E_(2) = 17`
`But E_(3) - E_(2) = (13.6)/(9)Z^(2) - ((-13.6)/(4)Z^(2))`
`= - 13.6 Z^(2) [(1)/(9) - (1)/(4)]`
`= - 13.6 Z^(2) [(4 - 9)/(36)] = (13.6 xx 5)/(36)Z^(2)`
` (13.6 xx 5)/(36)Z^(2) = 17 rArr Z= 3
`E_(n) - E_(2) = - (13.6)/(n^(2)) xx 3^(2) - [-(13.6)/(2^(2) xx 3^(2) ]`
`- 13.6 [(9)/(n^(2)) - (9)/(4)] = -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]` ....(ii)
frequency (i) and (ii)
` -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]= 27.2`
`rArr 122.4 (4 - n^(2)) = 108.8 n^(2)`
`rArr n^(2)= (489)/(13.6) = 36 rArr n = 6`
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