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 Relationship Between Total Energy and Kinetic Energy
In a hydrogen atom, the total energy (E) of the electron in a particular state is given by the sum of its kinetic energy (K.E.) and potential energy (P.E.). The relationship can be expressed as:
\[ E = K.E. + P.E. \]

### Step 2: Use the Known Values for the Ground State
For the ground state of the hydrogen atom, the total energy is given as:
\[ E = -13.6 \, \text{eV} \]
In the hydrogen atom, the potential energy (P.E.) is twice the negative of the kinetic energy (K.E.):
\[ P.E. = -2 \times K.E. \]
Thus, we can express the total energy in terms of kinetic energy:
\[ E = K.E. - 2 \times K.E. = -K.E. \]
From this, we can find the kinetic energy in the ground state:
\[ -K.E. = -13.6 \, \text{eV} \]
This implies:
\[ K.E. = 13.6 \, \text{eV} \]

### Step 3: Determine the Kinetic Energy in the First Excited State
In the first excited state (n=2), the total energy of the electron is given by:
\[ E_n = \frac{-13.6 \, \text{eV}}{n^2} \]
For 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 in the First Excited State
Using the relationship between total energy and kinetic energy:
\[ E = K.E. - 2 \times K.E. \]
We can rearrange this to find the kinetic energy:
\[ -K.E. = -3.4 \, \text{eV} \]
Thus:
\[ K.E. = 3.4 \, \text{eV} \]

### Final Answer
The kinetic energy of the electron in the first excited state of the hydrogen atom is:
\[ K.E. = 3.4 \, \text{eV} \]

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