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A hydrogen like atom (atomic number Z) i...

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition ot the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectivley Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

Text Solution

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From the given conditions, `E_n - E_2 = (10.2+17)eV = 27.2 eV` ….
and `E_n-E_3 = (4.25 +5.95) eV = 10.2 eV` ….(ii)
Eq (i) - Eq(ii) gives `E_3 - E_2 = 17.0 eV or Z^2 (13.6) ((1)/(4) - (1)/(9)) = 17.0`
`rArr Z^2 (13.6) (5//36) = 17.0`
`rArr Z^2 = 9 or Z=3`
From Eq .(i), `Z^2 (13.6) ((1)/(4) - (1)/(n^2)) = 27.2 or (3)^2 (13.6) ((1)/(4)- (1)(n^2) = 0.222 or 1/n^2 = 0.0278`
or ` n^2 =36 rArr n = 6.`
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