The total energy of eletcron in the ground state of hydrogen atom is `-13.6 eV`. The kinetic enegry of an electron in the first excited state is

Options

A. `1.7 eV`
B. `3.4 eV`
C. `6.8 eV`
D. `13.6 eV`

Answer

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

### Step 1: Understand the energy formula
According to Bohr's atomic model, the total energy \( E_n \) of an electron in the nth energy level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step 2: Determine the energy for the first excited state
The first excited state corresponds to \( n = 2 \). We can substitute \( n = 2 \) into the energy formula:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 3: Relate total energy to kinetic energy
In Bohr's model, the kinetic energy \( K \) of the electron in any orbit is related to the total energy \( E \) by the equation:
\[ K = -\frac{1}{2} E \]

### Step 4: Calculate the kinetic energy
Now, substituting the total energy \( E_2 = -3.4 \, \text{eV} \) into the kinetic energy formula:
\[ K = -\left(-\frac{1}{2} \times 3.4 \, \text{eV}\right) = \frac{3.4 \, \text{eV}}{2} = 3.4 \, \text{eV} \]

### Conclusion
Thus, the kinetic energy of the electron in the first excited state is:
\[ K = 3.4 \, \text{eV} \]