The orbital velocity of electron in the ground state is `v`. If the electron is excited to enegry staet `- 0.54 eV` its orbital velocity will be

Options

A. `v`
B. `(v)/(3)`
C. `(v)/(5)`
D. `(v)/(7)`

Explanation

To solve the problem, we need to find the orbital velocity of an electron when it is excited to an energy state of -0.54 eV, given that its orbital velocity in the ground state is \( v \).

### Step-by-Step Solution:

1. **Understanding the Ground State:**
The ground state of hydrogen-like atoms corresponds to the principal quantum number \( n = 1 \). The energy of the electron in the ground state can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For \( n = 1 \):
\[ E_1 = -13.6 \, \text{eV} \]

2. **Finding the Principal Quantum Number for -0.54 eV:**
We need to find the principal quantum number \( n \) when the energy is -0.54 eV:
\[ -0.54 = -\frac{13.6}{n^2} \]
Rearranging gives:
\[ n^2 = \frac{13.6}{0.54} \]
Calculating \( n^2 \):
\[ n^2 \approx 25.1852 \implies n \approx 5 \]

3. **Relating Orbital Velocity to Principal Quantum Number:**
The orbital velocity \( v_n \) of an electron in the nth orbit is inversely proportional to \( n \):
\[ v_n \propto \frac{1}{n} \]
Therefore, if the velocity in the ground state (n=1) is \( v \), then:
\[ v_n = \frac{v}{n} \]

4. **Calculating the Orbital Velocity in the Excited State:**
Since we found \( n = 5 \):
\[ v_5 = \frac{v}{5} \]

### Final Answer:
The orbital velocity of the electron when it is excited to the energy state of -0.54 eV will be:
\[ \boxed{\frac{v}{5}} \]