An electron beam has a kinetic energy equal to 100 eV . Find its wavelength associated with a beam , if mass of electron ` = 9.1 xx 10^(-31) " kg and 1 eV " = 1.6 xx 10^(-19) J ` . (Planks's constant = ` 6.6 xx 10^(-34) J-s)`

Options

A. `24.6 Å`
B. `0.12 Å`
C. `1.2 Å`
D. `6.3 Å`

Answer

To find the wavelength associated with an electron beam that has a kinetic energy of 100 eV, we can use the de Broglie wavelength formula. The formula for the wavelength (λ) of a particle is given by:

\[ \lambda = \frac{h}{\sqrt{2 m E}} \]

where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( E \) is the kinetic energy of the electron.

### Step-by-Step Solution:

**Step 1: Convert the kinetic energy from eV to joules.**

Given:
- Kinetic energy \( E = 100 \, \text{eV} \)
- \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)

So,
\[ E = 100 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-17} \, \text{J} \]

**Step 2: Substitute the values into the de Broglie wavelength formula.**

Given:
- Planck's constant \( h = 6.6 \times 10^{-34} \, \text{J s} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)

Now substituting these values into the formula:
\[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}} \]

**Step 3: Calculate the denominator.**

First, calculate \( 2 \times 9.1 \times 10^{-31} \):
\[ 2 \times 9.1 \times 10^{-31} = 1.82 \times 10^{-30} \]

Now multiply by \( 1.6 \times 10^{-17} \):
\[ 1.82 \times 10^{-30} \times 1.6 \times 10^{-17} = 2.912 \times 10^{-47} \]

Now take the square root:
\[ \sqrt{2.912 \times 10^{-47}} \approx 5.39 \times 10^{-24} \]

**Step 4: Calculate the wavelength.**

Now substitute back into the wavelength formula:
\[ \lambda = \frac{6.6 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.22 \times 10^{-10} \, \text{m} \]

**Step 5: Convert to angstroms.**

Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 1.22 \, \text{Å} \]

### Final Answer:
The wavelength associated with the electron beam is approximately \( 1.22 \, \text{Å} \).