A) - 54.4 eV
B) -27.2eV
C) - 6.8 eV
D) - 3.4 eV
Correct Answer: D
Solution :
Key Idea: The energy of hydrogen atom in its nth excited state is given by \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] For ground state \[(n=1)\], \[{{E}_{1}}=-\frac{13.6}{{{(1)}^{2}}}=-13.6eV\] For first excited state \[(n=2)\], \[{{E}_{2}}=-\frac{13.6}{{{(2)}^{2}}}=-\frac{13.6}{4}\] \[=-3.4eV\] NOTE: In ground state (n =1) energy of atom is -13.6 and energy corresponding to \[n=\infty \] is zero. Therefore, energy required to remove the electron from ground state is 13.6 eV.You need to login to perform this action.
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