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