A) 182.51 Å
B) 177.17 Å
C) 142.25 Å
D) 113.74 Å
Correct Answer: D
Solution :
\[{{E}_{n}}=-13.6\frac{{{Z}^{2}}}{{{n}^{2}}}eV.\] Required energy for said transition \[\Delta E={{E}_{3}}-{{E}_{1}}=13.6\ {{Z}^{2}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[\Rightarrow \Delta E=13.6\times {{3}^{2}}\left[ \frac{8}{9} \right]=108.8\ eV\] \[\Rightarrow \Delta E=108.8\times 1.6\times {{10}^{-19}}J\] Now \[\Delta E=\frac{hc}{\lambda }=108.8\times 1.6\times {{10}^{-19}}\] \[\Rightarrow \lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{108.8\times 1.6\times {{10}^{-19}}}=0.11374\times {{10}^{-7}}m\]\[=113.74{\AA}\]
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