In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, then

Options

A. `E=6.8eV,lambda=6.6xx10^(-10)m`
B. `E=3.4eV,lambda=6.6xx10^(-10)m`
C. `E=3.4eV,lambda=6.6xx10^(-11)m`
D. `E=6.8eV,lambda=6.6xx10^(-11)m`

Explanation

To solve the problem, we need to find the kinetic energy (E) of the electron and its de Broglie wavelength (λ) in an excited state of a hydrogen-like atom where the total energy is given as -3.4 eV.

### Step-by-Step Solution:

1. **Understanding Total Energy**:
The total energy (E_total) of the electron in an atom is given by the sum of its kinetic energy (E_k) and potential energy (E_p):
\[ E_{\text{total}} = E_k + E_p \]
In a hydrogen-like atom, the potential energy is related to the kinetic energy by:
\[ E_p = -2E_k \]

2. **Setting Up the Equation**:
Given that the total energy is -3.4 eV, we can substitute the potential energy into the total energy equation:
\[ E_{\text{total}} = E_k + (-2E_k) = -E_k \]
Therefore, we can write:
\[ -E_k = -3.4 \, \text{eV} \]

3. **Solving for Kinetic Energy**:
From the equation above, we can find the kinetic energy:
\[ E_k = 3.4 \, \text{eV} \]

4. **Calculating the de Broglie Wavelength**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy:
\[ p = \sqrt{2mE_k} \]
where \( m \) is the mass of the electron and \( E_k \) is the kinetic energy.

5. **Substituting Values**:
The Planck constant \( h \) is approximately \( 6.63 \times 10^{-34} \, \text{Js} \), and the mass of the electron \( m \) is approximately \( 9.1 \times 10^{-31} \, \text{kg} \). We need to convert the kinetic energy from eV to Joules:
\[ E_k = 3.4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 5.44 \times 10^{-19} \, \text{J} \]

6. **Calculating Momentum**:
Now we can calculate the momentum:
\[ p = \sqrt{2 \times (9.1 \times 10^{-31} \, \text{kg}) \times (5.44 \times 10^{-19} \, \text{J})} \]

7. **Finding the de Broglie Wavelength**:
Finally, substituting \( p \) back into the wavelength formula:
\[ \lambda = \frac{h}{p} \]

8. **Final Calculation**:
After calculating \( p \) and substituting it into the equation for \( \lambda \), we find:
\[ \lambda \approx 6.6 \times 10^{-10} \, \text{m} \]

### Summary of Results:
- Kinetic Energy \( E_k = 3.4 \, \text{eV} \)
- de Broglie Wavelength \( \lambda \approx 6.6 \times 10^{-10} \, \text{m} \)