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 the electron in a hydrogen-like atom with a total energy of -3.4 eV, we will follow these steps:

### Step 1: Determine the Kinetic Energy of the Electron
In a hydrogen-like atom, the total energy \( E \) of the electron is related to its kinetic energy \( KE \) by the equation:
\[ KE = -E \]
Given that the total energy \( E = -3.4 \, \text{eV} \), we can find the kinetic energy:
\[ KE = -(-3.4 \, \text{eV}) = 3.4 \, \text{eV} \]

### Step 2: Convert Kinetic Energy to Joules
We need to convert the kinetic energy from electron volts to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ KE = 3.4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 5.44 \times 10^{-19} \, \text{J} \]

### Step 3: Use the de-Broglie Wavelength Formula
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron. The momentum \( p \) can be expressed in terms of kinetic energy:
\[ p = \sqrt{2 m_{e} KE} \]
where \( m_{e} \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \, \text{kg} \).

### Step 4: Calculate the Momentum
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:
\[ p \approx \sqrt{9.88 \times 10^{-48}} \approx 3.14 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Calculate the de-Broglie Wavelength
Now substituting \( p \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \, \text{J s}}{3.14 \times 10^{-24} \, \text{kg m/s}} \approx 2.11 \times 10^{-10} \, \text{m} \]
Converting this to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 2.11 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of the electron is approximately \( 2.11 \, \text{Å} \).

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