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