An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV, find the de-Broglie wavelength of the electron.

Explanation

To find the de Broglie wavelength of an electron in a hydrogen-like atom with a total energy of -3.4 eV, we can follow these steps:

### Step 1: Convert the total energy from eV to Joules
The total energy \( E \) of the electron is given as -3.4 eV. To convert this energy into Joules, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \).

\[ E = -3.4 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = -5.44 \times 10^{-19} \text{ J} \]

### Step 2: Calculate the kinetic energy
In a hydrogen-like atom, the total energy \( E \) is related to the kinetic energy \( K \) by the equation:

\[ K = -E \]

Thus, the kinetic energy of the electron is:

\[ K = 5.44 \times 10^{-19} \text{ J} \]

### Step 3: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) 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:

\[ p = \sqrt{2mK} \]

where \( m \) is the mass of the electron (\( m = 9.1 \times 10^{-31} \text{ kg} \)).

### Step 4: Substitute the values to find momentum
Now, substituting the values into the momentum equation:

\[ p = \sqrt{2 \times (9.1 \times 10^{-31} \text{ kg}) \times (5.44 \times 10^{-19} \text{ J})} \]

Calculating this gives:

\[ p = \sqrt{9.9 \times 10^{-49}} \approx 9.95 \times 10^{-25} \text{ kg m/s} \]

### Step 5: Calculate the de Broglie wavelength
Now we can substitute \( p \) into the de Broglie wavelength formula. The Planck's constant \( h \) is \( 6.63 \times 10^{-34} \text{ Js} \).

\[ \lambda = \frac{6.63 \times 10^{-34} \text{ Js}}{9.95 \times 10^{-25} \text{ kg m/s}} \approx 6.66 \times 10^{-10} \text{ m} \]

### Step 6: Convert the wavelength to nanometers
To convert meters to nanometers, we multiply by \( 10^9 \):

\[ \lambda \approx 0.666 \text{ nm} \]

### Final Answer
The de Broglie wavelength of the electron is approximately \( 0.66 \text{ nm} \).

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