In a hydrogen atom, the energy of the electron in the $n$th orbit is $E_n = -\tfrac{13.6}{n^2}\,\text{eV}$. The energy required to excite the electron from $n = 1$ to $n = 3$ is
A$12.09\,\text{eV}$
B$10.2\,\text{eV}$
C$1.51\,\text{eV}$
D$13.6\,\text{eV}$
Answer & Solution
Correct answer: A. $12.09\,\text{eV}$
$\Delta E = E_3 - E_1 = -\tfrac{13.6}{9} - (-13.6) = 13.6\left(1 - \tfrac{1}{9}\right) = 13.6 \cdot \tfrac{8}{9} \approx 12.09\,\text{eV}$. Option A is the $1\to 2$ jump (10.2 eV); option C is the full ionisation energy from ground state.
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