A) 1.51
B) 13.6
C) 40.8
D) 122.4
Correct Answer: D
Solution :
[d] \[{{E}_{n}}=-13.6\frac{{{\left( Z \right)}^{2}}}{\left( {{n}^{2}} \right)}eV\] Therefore, ground state energy of double ionized lithium atom (Z=3, n=1) will be \[{{E}_{1}}=\left( -13.6 \right)\frac{{{\left( 3 \right)}^{2}}}{{{\left( 1 \right)}^{2}}}=-122.4eV\] \[\therefore \]Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4eV
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