A) two
B) three
C) four
D) one
Correct Answer: B
Solution :
Ionization energy corresponding to ionization potential \[=-13.6\,eV\] Photon energy incident = 121 eV So, the energy of electron in excited state \[=-13.6+12.1=-1.5\,eV\] \[\Rightarrow \]\[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV-1.5=\frac{-13.6}{{{n}^{2}}}\] \[{{n}^{2}}=\frac{-13.6}{-1.5}=9\] \[\therefore \] \[n=3\] This is energy of electron in excited state corresponds to third orbit. The possible spectral lies are when electron jumps form orbit 3rd to 2nd; to 1st and 2nd to 1st.Thus, 3 spectral lines are emitted.You need to login to perform this action.
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