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Question

If the ionization energy of hydrogen atom is 13.6eV then the wavelength of the radiation required to excite the electron in Li++ from first to third Bohr orbit is approximately

A
1140A
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B
914A
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C
11.4A
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D
134A
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Solution

The correct option is B 914A
Ionisation energy is given by
E=13.6eV (given)
or E=13.6×1.6×1019J ....(1)
also E=hv
E=hcλ ....(2)
equation (1) and (2)
hcλ=13.6×1.6×1019
λ=h×c13.6×1.6×1019
λ=3×108×6.63×103413.6×1.6×1019
λ=914×1010m
λ=914Ao

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