A) 6.04 eV
B) 13.60 eV
C) 54.40 eV
D) 48.36 eV
Correct Answer: B
Solution :
Energy levels in Hydrogen like atom is given by \[E=-13.6\frac{{{z}^{2}}}{{{n}^{2}}}eV\] As \[H{{e}^{+}}\] is 1st excited state \[\therefore \]\[z=2,n=2\] \[E=-13.6eV\] As total energy of \[H{{e}^{+}}\]in 1st excited state is \[13.6\text{ }eV,\]ionisation energy should be \[+13.6eV.\]You need to login to perform this action.
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