A) \[0.563\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[4861\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[4102\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[1213\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
For ionization energy of electron in nth orbit \[{{E}_{n}}=\frac{-13.6}{{{n}^{2}}}eV\] For an electron in Ist orbit, i.e.,\[n=1\] \[{{E}_{1}}=\frac{-13.6}{{{1}^{2}}}=-13.6\,eV\] For an electron in \[\text{II}\,\text{nd}\]orbit, i.e.,\[~n=2\] \[{{E}_{2}}=\frac{-13.6}{{{2}^{2}}}=-3.4\,eV\] Therefore, energy released \[\Delta E={{E}_{2}}-{{E}_{1e}}\] \[=(-3.4)-(-13.6)\] \[=10.2\,eV\] Therefore, wavelength of line emitted is given by \[\lambda =\frac{12375}{\Delta E}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=\frac{12375}{10.2}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=1213\overset{\text{o}}{\mathop{\text{A}}}\,\]
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