A) -30.6 eV
B) 27.2 eV
C) -13.6 eV
D) - 27.2 eV
Correct Answer: A
Solution :
[a] Energy of electron in nth orbit is \[{{E}_{n}}=-\left( Rch \right)\frac{{{Z}^{2}}}{{{n}^{2}}}=-54.4eV\] For \[H{{e}^{+}}\] is ground state \[{{E}_{1}}=-(Rch)\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=-\,54.4\,\Rightarrow \,Rch\,=13.6\] \[\therefore \] For \[L{{i}^{++}}\] in first excited state (n=2) \[E=-13.6\times \frac{{{\left( 3 \right)}^{2}}}{{{\left( 2 \right)}^{2}}}=-30.6eV\]You need to login to perform this action.
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