The wavelength of an electron of energy 10 keV will be :

Options

A. `0.12 Å`
B. `1.2 Å`
C. `12 Å`
D. `120 Å`

Answer

To find the wavelength of an electron with an energy of 10 keV, we can use the de Broglie wavelength formula, which relates the wavelength (λ) to the momentum (P) of a particle. The steps to solve the problem are as follows:

### Step 1: Convert the energy from keV to joules
The energy given is 10 keV. We need to convert this energy into joules for our calculations.

\[ E = 10 \, \text{keV} = 10 \times 10^3 \, \text{eV} = 10 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J} \]

Calculating this gives:

\[ E = 10 \times 10^3 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-15} \, \text{J} \]

### Step 2: Calculate the momentum (P) of the electron
The kinetic energy (E) of the electron is related to its momentum (P) by the formula:

\[ E = \frac{P^2}{2m} \]

Rearranging this gives:

\[ P = \sqrt{2mE} \]

Where \( m \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \, \text{kg} \).

Substituting the values:

\[ P = \sqrt{2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-15})} \]

Calculating this gives:

\[ P = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-15}} \approx \sqrt{2.912 \times 10^{-45}} \approx 5.39 \times 10^{-23} \, \text{kg m/s} \]

### Step 3: Calculate the wavelength (λ) using de Broglie’s formula
The de Broglie wavelength is given by:

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

Where \( h \) is Planck's constant, \( h \approx 6.626 \times 10^{-34} \, \text{Js} \).

Substituting the values:

\[ \lambda = \frac{6.626 \times 10^{-34}}{5.39 \times 10^{-23}} \]

Calculating this gives:

\[ \lambda \approx 1.23 \times 10^{-11} \, \text{m} \]

### Final Answer
The wavelength of an electron with an energy of 10 keV is approximately:

\[ \lambda \approx 1.23 \times 10^{-11} \, \text{m} \text{ or } 0.123 \, \text{nm} \]

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