Question

The ground state energy of hydrogen atom is -13.6 eV. The energy of the second excited state is

• A. -3.4 eV
• B. -6.8 eV
• C. -1.51 eV
• D. -27.2 eV
• E. -0.85 eV

Hint:

Be extremely careful with the wording! Students often confuse "second excited state" with $n=2$. Always remember: $n = (\text{Excited State Number}) + 1$.

Answer 1

The Correct Option is C

Concept:
Physics - Bohr's Atomic Model.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = \frac{E_1}{n^2}$, where $E_1$ is the ground state energy.
Step 1: Identify the principal quantum number (n).

• Ground state: $n = 1$
• First excited state: $n = 2$
• Second excited state: $n = 3$

Step 2: Apply the energy formula.
Given $E_1 = -13.6$ eV and $n = 3$: $$ E_3 = \frac{-13.6}{3^2} $$
Step 3: Calculate the final value.
$$ E_3 = \frac{-13.6}{9} \approx -1.51 \text{ eV} $$