A doubly ionized lithium atom is hydrogen like with atomic. number 3. Find the wavelength of the radiation to excite the electron in. Li++ form the first to the third Bohr orbit. The ionization energy of the hydrogen. Atom is 13.6V.
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The electron in the Li++ ion is excited from the first bohr orbit to the third bohr orbit.
We will find the energy for this electron excitation.
E = 13.6 × Z² ((1/(n1)²) - (1/(n2)²)
Put the values Z = 3 , n1 = 1 , n2 = 3
=> E = 13.6 × 9 (1 - (1/9))
=> E = 108.8 eV
Wavelength of the radiation in A° = 12420 / Energy of the radiation in eV
= 12420/108.8
= 114.15 A°
The wavelength of the radiation emitted from the excitation of the electron = 114.15 A°
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