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. `lambda=6.6Å`
B. `E=3.4eV`
C. Both are correct
D. Both are wrong

Explanation

To solve the problem step by step, we will analyze the given information and apply relevant physics concepts.

### Step 1: Understand the Total Energy of the Electron
In a hydrogen-like atom, the total energy (E) of an electron in an excited state is given as:
\[ E = -3.4 \, \text{eV} \]

### Step 2: Relate Total Energy to Kinetic Energy
For an electron in a bound state, the total energy (E) is related to its kinetic energy (K) and potential energy (U) as follows:
\[ E = K + U \]
In a hydrogen-like atom, the potential energy (U) is twice the negative of the kinetic energy (K):
\[ U = -2K \]
Thus, we can express the total energy as:
\[ E = K - 2K = -K \]
From this, we can deduce that:
\[ K = -E \]
Substituting the value of E:
\[ K = -(-3.4 \, \text{eV}) = 3.4 \, \text{eV} \]

### Step 3: Calculate the de-Broglie Wavelength
The de-Broglie wavelength (\( \lambda \)) of an electron 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 (K):
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the electron.

Substituting for \( p \):
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

### Step 4: Substitute Known Values
Using the known values:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Kinetic energy \( K = 3.4 \, \text{eV} = 3.4 \times 1.6 \times 10^{-19} \, \text{J} \)

Now substituting these values into the equation for \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times (3.4 \times 1.6 \times 10^{-19})}} \]

### Step 5: Calculate the Value of \( \lambda \)
Calculating the value inside the square root:
\[ 2 \times 9.11 \times 10^{-31} \times (3.4 \times 1.6 \times 10^{-19}) \]
\[ = 2 \times 9.11 \times 10^{-31} \times 5.44 \times 10^{-19} \]
\[ = 9.88 \times 10^{-48} \]

Now calculating \( \sqrt{9.88 \times 10^{-48}} \):
\[ \sqrt{9.88 \times 10^{-48}} \approx 3.14 \times 10^{-24} \]

Now substituting back to find \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{3.14 \times 10^{-24}} \approx 2.11 \times 10^{-10} \, \text{m} = 2.11 \, \text{Å} \]

### Conclusion
From the calculations, we find:
1. The kinetic energy \( E \) of the electron is \( 3.4 \, \text{eV} \).
2. The de-Broglie wavelength \( \lambda \) is approximately \( 2.11 \, \text{Å} \).

### Final Answer
- Kinetic Energy \( E = 3.4 \, \text{eV} \)
- de-Broglie Wavelength \( \lambda \approx 2.11 \, \text{Å} \)