Calculate the de Broglie wavelength for a beam of electron whose energy is `100 eV`

Explanation

To calculate the de Broglie wavelength for a beam of electrons with an energy of 100 eV, we can follow these steps:

### Step 1: Convert the Energy from eV to Joules
The energy of the electron beam is given as 100 eV. To convert this energy into joules, we use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
So,
\[ E = 100 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.6 \times 10^{-17} \text{ J} \]

### Step 2: Relate Energy to Kinetic Energy
The kinetic energy (KE) of the electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity. We can rearrange this to find the velocity:
\[ v = \sqrt{\frac{2 \times KE}{m}} \]

### Step 3: Substitute the Mass of the Electron
The mass of the electron is approximately:
\[ m = 9.1 \times 10^{-31} \text{ kg} \]
Now substituting the values:
\[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-17}}{9.1 \times 10^{-31}}} \]

### Step 4: Calculate the Velocity
Calculating the above expression:
\[ v = \sqrt{\frac{3.2 \times 10^{-17}}{9.1 \times 10^{-31}}} = \sqrt{3.51 \times 10^{13}} \approx 5.93 \times 10^6 \text{ m/s} \]

### Step 5: Calculate the Momentum
The momentum \( p \) of the electron can be calculated using:
\[ p = mv \]
Substituting the values we have:
\[ p = 9.1 \times 10^{-31} \text{ kg} \times 5.93 \times 10^6 \text{ m/s} \approx 5.39 \times 10^{-24} \text{ kg m/s} \]

### Step 6: Calculate the de Broglie Wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant, approximately:
\[ h = 6.626 \times 10^{-34} \text{ J s} \]
Substituting the values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.23 \times 10^{-10} \text{ m} \]

### Step 7: Convert to Angstroms
Since \( 1 \text{ Å} = 10^{-10} \text{ m} \), we can express the wavelength in angstroms:
\[ \lambda \approx 1.23 \text{ Å} \]

### Final Answer
The de Broglie wavelength of the electron beam is approximately \( 1.23 \text{ Å} \).

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