In a sample of hydrogen atom in ground state electrons make transition from ground state to a particular excited state where path length is five times de-broglie wavelength, electrons make back transition to the ground state producing all possible photons. If photon having 2nd highest energy of this sample can be used to excite the electron in a particular excited state of Li 2+ atom then find the final excited state of Li 2+ atom.
Text Solution
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In the particular excited state of H-atom, path length is five times the de-Broglie wavelength.
∴ 2 π r = 5 λ ....
However, path length in a state n is n times the de-Broglie wavelength.
∴ 2 π r = n λ ....
From & , principal quantum number (n) of the excited state = 5. Photon having 2 nd highest energy corresponds to back transition of electron from n = 4 to n = 1
This photon will cause an already excited Li 2+ electron to go to some higher state. Let the initial excited state of Li 2+ ion be n 1 and final excited state of Li 2+ ion be n 2
∴ 
= 
∴ 
or 
On comparing both sides,
= 1 & 
= 4 
n 1 = 3 & n 2 = 12
Thus, the final excited state of Li 2+ ion electron is n = 12 Ans.
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