In a hydrogen atom, an electron is excited to the energy state of -1.511 eV. What will be the speed of the electron in orbit, if v is the speed in its ground state?

Options

A. Twice the speed of electron in ground state
B. Twice the speed of electron in ground state
C. The orbital speed will remain same
D. One-third of the speed of electron in ground state.

Explanation

To solve the problem, we need to determine the speed of the electron in the third energy state of a hydrogen atom, given that the speed in the ground state (first energy level) is denoted as \( v \).

### Step-by-Step Solution:

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

2. **Determine the Principal Quantum Number (n)**:
We are given that the energy of the electron in the excited state is \( -1.511 \, \text{eV} \). We can set up the equation:
\[ -1.511 = -\frac{13.6}{n^2} \]
Rearranging gives:
\[ n^2 = \frac{13.6}{1.511} \]
Calculating this:
\[ n^2 \approx 9 \quad \Rightarrow \quad n \approx 3 \]
Thus, the electron is in the third orbit (n=3).

3. **Use the Velocity Formula**:
The velocity of an electron in the nth orbit is given by:
\[ v_n = v_0 \cdot \frac{Z}{n} \]
where \( v_0 \) is the speed in the ground state (n=1), and \( Z \) is the atomic number. For hydrogen, \( Z = 1 \).

4. **Calculate the Speed in the Third Orbit**:
Since \( Z = 1 \) for hydrogen, we can substitute into the formula:
\[ v_3 = v_0 \cdot \frac{1}{3} \]
Since \( v_0 = v \) (the speed in the ground state), we have:
\[ v_3 = \frac{v}{3} \]

5. **Final Answer**:
Therefore, the speed of the electron in the third energy state is:
\[ v_3 = \frac{v}{3} \]

### Summary:
The speed of the electron in the excited state (n=3) is one-third of the speed in the ground state (n=1).