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. Here’s a step-by-step solution:

### Step 1: Convert Kinetic Energy from eV to Joules
The kinetic energy (KE) of the electron is given as 100 eV. We need to convert this energy into joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ KE = 100 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.6 \times 10^{-17} \text{ J} \]

### Step 2: Use the Kinetic Energy to Find Momentum
The kinetic energy of a particle can also be expressed in terms of momentum \(p\) using the formula:

\[ KE = \frac{p^2}{2m} \]

Rearranging this formula to solve for momentum \(p\):

\[ p = \sqrt{2m \cdot KE} \]

Substituting the mass of the electron \(m = 9.1 \times 10^{-31} \text{ kg}\) and the kinetic energy we calculated:

\[ p = \sqrt{2 \times (9.1 \times 10^{-31} \text{ kg}) \times (1.6 \times 10^{-17} \text{ J})} \]

Calculating this gives:

\[ p = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}} = \sqrt{2.912 \times 10^{-47}} \approx 5.39 \times 10^{-24} \text{ kg m/s} \]

### Step 3: Calculate the Wavelength Using de Broglie Relation
The de Broglie wavelength \(\lambda\) is given by the formula:

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

Where \(h\) is Planck's constant, \(h = 6.6 \times 10^{-34} \text{ J s}\). Substituting the values we have:

\[ \lambda = \frac{6.6 \times 10^{-34} \text{ J s}}{5.39 \times 10^{-24} \text{ kg m/s}} \]

Calculating this gives:

\[ \lambda \approx 1.22 \times 10^{-10} \text{ m} \]

### Step 4: Convert Wavelength to Angstroms
Since \(1 \text{ angstrom} = 10^{-10} \text{ m}\):

\[ \lambda \approx 1.22 \text{ angstroms} \]

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

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