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