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Question

# An energy of 68.0eV is required to excite a hydrogen-like atom in its second Bohr energy level to third energy level the charge of nucleus is Ze. The wavelength of a radiation required to eject the electron from first orbit to infinity is

A
2.2nm
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B
2.85nm
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C
3.2nm
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D
2.5nm
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Solution

## The correct option is D 2.5nmGiven, n1=2, n2=3 and ΔE=68eVThe difference in energies of two orbitsΔE=13.6Z2[1n21−1n22]where n1<n268=13.6Z2[122−132] =13.6Z2[14−19] =13.6Z2[9−436] =13.6Z2×536∴Z2=68×3613.6×5=36Z=6∵ Wavelength of photon, 1λ=Z2R[1n21−1n22]n1=1 and n2=∞1λ=62×R[112−1∞]=36R∴λ=136R=136×1.097×107 =10−739.5m =0.025×10−7m =2.5nm

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