A) 3.4 eV
B) 13.6 eV
C) 54.4 eV
D) 122.4 eV
Correct Answer: C
Solution :
For third line of Balmer series \[{{n}_{1}}=2\], \[{{n}_{2}}=5\] \ \[\frac{1}{\lambda }=R{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] gives \[{{Z}^{2}}=\frac{n_{1}^{2}n_{2}^{2}}{(n_{2}^{2}-n_{1}^{2})\lambda R}\] On putting values Z = 2 From \[E=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}=\frac{-13.6{{(2)}^{2}}}{{{(1)}^{2}}}=-54.4\,eV\]
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