What must be the velocity of a beam of electron if they are to display a de Broglie wavelength of `1 Å`

Explanation

To find the velocity of a beam of electrons that displays a de Broglie wavelength of \(1 \, \text{Å}\) (angstrom), we will use the de Broglie wavelength formula:

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

Where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is Planck's constant (\(6.625 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \(v\) is the velocity of the electron.

### Step-by-Step Solution:

1. **Convert the Wavelength**:
The given wavelength is \(1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}\).

2. **Rearrange the de Broglie Equation**:
We need to find \(v\), so we rearrange the equation:
\[ v = \frac{h}{m\lambda} \]

3. **Substitute the Values**:
Substitute \(h\), \(m\), and \(\lambda\) into the equation:
\[ v = \frac{6.625 \times 10^{-34} \, \text{Js}}{(9.1 \times 10^{-31} \, \text{kg})(1 \times 10^{-10} \, \text{m})} \]

4. **Calculate the Denominator**:
First, calculate the product of mass and wavelength:
\[ 9.1 \times 10^{-31} \, \text{kg} \times 1 \times 10^{-10} \, \text{m} = 9.1 \times 10^{-41} \, \text{kg m} \]

5. **Calculate the Velocity**:
Now, substitute this back into the equation for \(v\):
\[ v = \frac{6.625 \times 10^{-34}}{9.1 \times 10^{-41}} \]
\[ v \approx 7.29 \times 10^{6} \, \text{m/s} \]

### Final Answer:
The velocity of the beam of electrons must be approximately \(7.29 \times 10^{6} \, \text{m/s}\).

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