Question:
The electron in the hydrogen atom is in the third excited state. Its potential energy (in eV) is
The electron in the hydrogen atom is in the third excited state. Its potential energy (in eV) is
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Remember that:
$\text{Total Energy (E)} = \text{Kinetic Energy (K)} + \text{Potential Energy (U)}$.
By the virial theorem for a $1/r$ potential, $K = -E$ and $U = 2E$.
$\text{Total Energy (E)} = \text{Kinetic Energy (K)} + \text{Potential Energy (U)}$.
By the virial theorem for a $1/r$ potential, $K = -E$ and $U = 2E$.
Updated On: Jun 16, 2026
- $-1.70$
- $-1.51$
- $-0.85$
- $-3.02$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem asks for the potential energy of an electron in a hydrogen atom when it is in its third excited state.
Step 2: Key Formula or Approach:
- The total energy $E_n$ of an electron in the $n$-th orbit of a hydrogen atom is:
\[ E_n = -\frac{13.6}{n^2}\text{ eV} \]
- For a Coulombic potential, the potential energy $U_n$ is related to the total energy $E_n$ by:
\[ U_n = 2E_n \]
Step 3: Detailed Explanation:
• The "third excited state" corresponds to the principal quantum number $n = 4$.
(Ground state is $n=1$, first excited state is $n=2$, second excited state is $n=3$, third excited state is $n=4$).
• Let us calculate the total energy $E_4$ in the $n = 4$ state:
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85\text{ eV} \]
• Using the relationship between potential energy and total energy:
\[ U_4 = 2E_4 = 2 \times (-0.85\text{ eV}) = -1.70\text{ eV} \]
Step 4: Final Answer:
Its potential energy is $-1.70$ eV.
The problem asks for the potential energy of an electron in a hydrogen atom when it is in its third excited state.
Step 2: Key Formula or Approach:
- The total energy $E_n$ of an electron in the $n$-th orbit of a hydrogen atom is:
\[ E_n = -\frac{13.6}{n^2}\text{ eV} \]
- For a Coulombic potential, the potential energy $U_n$ is related to the total energy $E_n$ by:
\[ U_n = 2E_n \]
Step 3: Detailed Explanation:
• The "third excited state" corresponds to the principal quantum number $n = 4$.
(Ground state is $n=1$, first excited state is $n=2$, second excited state is $n=3$, third excited state is $n=4$).
• Let us calculate the total energy $E_4$ in the $n = 4$ state:
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85\text{ eV} \]
• Using the relationship between potential energy and total energy:
\[ U_4 = 2E_4 = 2 \times (-0.85\text{ eV}) = -1.70\text{ eV} \]
Step 4: Final Answer:
Its potential energy is $-1.70$ eV.
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