A) 3.4eV
B) 3.6eV
C) 10.2eV
D) 13.6eV
Correct Answer: C
Solution :
The energy associated with first orbit of the hydrogen atom is given by \[{{E}_{1}}=\frac{13.6}{{{n}^{2}}}=-\frac{13.6}{{{1}^{2}}}=-13.6\,eV\] Energy associated with second orbit of hydrogen atom is given by \[{{E}_{2}}=\frac{13.6}{{{(2)}^{2}}}=-\frac{13.6}{4}=-13.4\,eV\] The minimum accelerating potential required to energise an electron which, on collision can excite an atom is called the excitation potential of that atom. Therefore, minimum excitation potential of the Bohrs first orbit in the hydrogen atom is given by \[E={{E}_{2}}-{{E}_{1}}=-3.4-(-13.6)=10.2\,eV\].
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