Question

Energy of electron of hydrogen atom in $3^{\text{rd}}$ excited state is

Hint:

Be very careful with terminology: the $n^{\text{th}}$ excited state always strictly corresponds to principal quantum number $n+1$.
Memorize the first four energy levels for hydrogen to save time: $-13.6\text{ eV}$, $-3.4\text{ eV}$, $-1.51\text{ eV}$, and $-0.85\text{ eV}$.

Answer 1

Step 1: Understanding the Concept:
In Bohr's highly successful model of the hydrogen atom, electrons occupy specific, discrete, and quantized circular orbits (energy levels).
The lowest possible energy level is correctly referred to as the ground state, while the higher energy levels above it are categorized sequentially as excited states.
Step 2: Key Formula or Approach:
The total energy $E_n$ of an electron orbiting in the $n^{\text{th}}$ principal quantum state of a hydrogen atom is calculated using the established formula:
\[ E_n = -\frac{13.6}{n^2}\text{ eV} \] Step 3: Detailed Explanation:
First, we must correctly identify the principal quantum number $n$ corresponding to the "3$^{\text{rd}}$ excited state."
The ground state is inherently $n = 1$.
The $1^{\text{st}}$ excited state corresponds to $n = 2$.
The $2^{\text{nd}}$ excited state corresponds to $n = 3$.
Therefore, the $3^{\text{rd}}$ excited state strictly corresponds to $n = 4$.
Now, substitute $n = 4$ directly into the quantized energy formula:
\[ E_4 = -\frac{13.6}{4^2}\text{ eV} \] \[ E_4 = -\frac{13.6}{16}\text{ eV} \] Executing the division yields the numerical value:
\[ E_4 = -0.85\text{ eV} \] Step 4: Final Answer:
The energy of the electron in the $3^{\text{rd}}$ excited state is $-0.85\text{ eV}$.