Total energy of an electron in the ground state of hydrogen atom is – 136 eV. Its total energy, when hydrogen atom is in the first excited state, is :

Options

A. `+ 13.6 eV `
B. `+ 3.4 eV`
C. `-3.4 eV `
D. `-54.4 eV `

Explanation

To find the total energy of an electron in the first excited state of a hydrogen atom, we can use the formula for the energy levels of the hydrogen atom:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

where \( E_n \) is the energy of the electron in the nth orbit, and \( n \) is the principal quantum number.

### Step 1: Identify the ground state energy
The total energy of an electron in the ground state (n=1) is given as:

\[ E_1 = -13.6 \, \text{eV} \]

### Step 2: Determine the first excited state
The first excited state corresponds to \( n = 2 \).

### Step 3: Calculate the energy for the first excited state
Using the formula for the energy levels, we substitute \( n = 2 \):

\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} \]

### Step 4: Simplify the calculation
Calculating \( 2^2 \):

\[ 2^2 = 4 \]

Now substitute this back into the equation:

\[ E_2 = -\frac{13.6 \, \text{eV}}{4} \]

### Step 5: Perform the division
Now, divide \( -13.6 \, \text{eV} \) by \( 4 \):

\[ E_2 = -3.4 \, \text{eV} \]

### Conclusion
The total energy of the electron in the first excited state of the hydrogen atom is:

\[ E_2 = -3.4 \, \text{eV} \]