Which transition of Li2+ is associated with same energy change as n=6 to n=4 transition in He+
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The correct answer is - n = 9 to n = 6
This is because the energies of the 2nd, 4th, and the 6th orbits in He⁺ are equal to the energies of the 3rd, 6th, and the 9th orbits in Li²⁺ respectively.
v⁻He⁺ = RZ² [ 1/4² - 1/6²]
= 4R [ 36 - 16/36 x 16] = 5R/36
5R/36 = R/4 - R/n2
On solving above equation
n² = 9
Thus, n = 3
The corresponding transition from 9 → 6 in Balmer series of Lithium atom has the same frequency as that of 6 → 4 transitions in He+.
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