If energy of electron in ground state is `-13.6` then find out speed of electron in fourth orbit of H-atom

Explanation

To find the speed of the electron in the fourth orbit of a hydrogen atom, we can use the formula for the speed of an electron in a stationary orbit, which is derived from the Bohr model of the hydrogen atom.

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number (orbit number).

2. **Finding the Energy in the Fourth Orbit**:
For the fourth orbit (\( n = 4 \)):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

3. **Using the Formula for Speed**:
The speed of the electron in the nth orbit can be calculated using the formula:
\[ v_n = \sqrt{-\frac{2E_n}{m}} \]
where \( E_n \) is the energy of the electron in the nth orbit and \( m \) is the mass of the electron (\( m \approx 9.11 \times 10^{-31} \, \text{kg} \)).

4. **Converting Energy to Joules**:
We need to convert the energy from electron volts to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ E_4 = -0.85 \, \text{eV} = -0.85 \times 1.6 \times 10^{-19} \, \text{J} = -1.36 \times 10^{-19} \, \text{J} \]

5. **Calculating the Speed**:
Now substituting the values into the speed formula:
\[ v_4 = \sqrt{-\frac{2 \times (-1.36 \times 10^{-19} \, \text{J})}{9.11 \times 10^{-31} \, \text{kg}}} \]
\[ v_4 = \sqrt{\frac{2.72 \times 10^{-19}}{9.11 \times 10^{-31}}} \]
\[ v_4 = \sqrt{2.98 \times 10^{11}} \approx 5.47 \times 10^5 \, \text{m/s} \]

### Final Answer:
The speed of the electron in the fourth orbit of the hydrogen atom is approximately \( 5.47 \times 10^5 \, \text{m/s} \).