The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. Which of the following is its kinetic energy in the first excited state?

Options

A. 13.6 eV
B. 6.8 eV
C. 3.4 eV
D. 1.825 eV

Answer

To find the kinetic energy of the electron in the first excited state of a hydrogen atom, we can follow these steps:

### Step 1: Understand the total energy in the ground state
The total energy \( E \) of the electron in the ground state (n=1) of a hydrogen atom is given as:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 2: Relate total energy to kinetic and potential energy
In a hydrogen atom, the total energy \( E \) is related to the kinetic energy \( K \) and potential energy \( U \) as follows:
\[ E = K + U \]
The potential energy \( U \) is twice the negative of the kinetic energy:
\[ U = -2K \]
Thus, we can rewrite the total energy as:
\[ E = K - 2K = -K \]
From this, we can find the kinetic energy in the ground state:
\[ K_1 = -E_1 = 13.6 \, \text{eV} \]

### Step 3: Determine the kinetic energy in the first excited state
For the first excited state (n=2), the total energy can be calculated using the formula:
\[ E_n = \frac{E_1}{n^2} \]
Substituting \( n = 2 \):
\[ E_2 = \frac{-13.6 \, \text{eV}}{2^2} = \frac{-13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 4: Calculate the kinetic energy for n=2
Using the relationship between total energy and kinetic energy:
\[ K_2 = -E_2 = 3.4 \, \text{eV} \]

### Final Answer
Thus, the kinetic energy of the electron in the first excited state (n=2) is:
\[ \boxed{3.4 \, \text{eV}} \]