If the first excitation energy for an H-like (hypothetical) sample is 24 eV, then the separation energy of an electron in the 3rd excited state is:
A
5 eV
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B
4 eV
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C
3 eV
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D
2 eV
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Solution
The correct option is D 2 eV First excitation energy n1→n2 ΔE1=RH[112−122]z2=34z2RH
Seperation energy from 3 rd excited state (n4→∞) E1=RH[142−1∞]z2=116z2RH
Taking ratio of E1andE2 we get, ΔE1ΔE2=34z2RH116z2RH
Given ΔE1=24eV ΔE2=ΔE112=24eV12=2eV