What must be the velocity of a beam of electrons if they are to display a de-Broglie wavelength of `100 Å` ?

Explanation

To find the velocity of a beam of electrons that displays a de-Broglie wavelength of \(100 \, \text{Å}\), we can use the de-Broglie wavelength formula:

\[ \lambda = \frac{h}{mv} \]

Where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is Planck's constant,
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron.

### Step 1: Convert the wavelength from angstroms to meters
Given:
\[ \lambda = 100 \, \text{Å} = 100 \times 10^{-10} \, \text{m} = 1.0 \times 10^{-8} \, \text{m} \]

### Step 2: Write down the values of constants
- Planck's constant \(h = 6.625 \times 10^{-34} \, \text{J s}\)
- Mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)

### Step 3: Rearrange the de-Broglie equation to solve for velocity \(v\)
From the de-Broglie equation:
\[ v = \frac{h}{m\lambda} \]

### Step 4: Substitute the known values into the equation
Substituting the values of \(h\), \(m\), and \(\lambda\):
\[ v = \frac{6.625 \times 10^{-34}}{(9.1 \times 10^{-31})(1.0 \times 10^{-8})} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ 9.1 \times 10^{-31} \times 1.0 \times 10^{-8} = 9.1 \times 10^{-39} \]

### Step 6: Calculate the velocity
Now substituting back into the equation:
\[ v = \frac{6.625 \times 10^{-34}}{9.1 \times 10^{-39}} = 7.28 \times 10^{4} \, \text{m/s} \]

### Final Answer
The velocity of the beam of electrons must be approximately:
\[ v \approx 7.28 \times 10^{4} \, \text{m/s} \]

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