The wavelenght of an electron of energy 10 keV will be

Options

A. 1.2Å
B. 0.12 Å
C. 12Å
D. 120Å

Explanation

To find the wavelength of an electron with an energy of 10 keV, we can follow these steps:

### Step 1: Convert Energy from keV to Joules
The given energy is 10 keV. We need to convert this into joules.

\[ \text{Energy (E)} = 10 \, \text{keV} = 10 \times 1000 \, \text{eV} = 10^4 \, \text{eV} \]

Using the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):

\[ E = 10^4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.6 \times 10^{-15} \, \text{J} \]

### Step 2: Use the Kinetic Energy Formula
The kinetic energy (KE) of the electron can be expressed in terms of momentum (p):

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

From this, we can solve for momentum:

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

### Step 3: Substitute Values
The mass (m) of an electron is approximately \(9.1 \times 10^{-31} \, \text{kg}\). Now substituting the values:

\[ p = \sqrt{2 \cdot (9.1 \times 10^{-31} \, \text{kg}) \cdot (1.6 \times 10^{-15} \, \text{J})} \]

Calculating this gives:

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

### Step 4: Calculate Wavelength Using de Broglie Relation
The de Broglie wavelength (\(\lambda\)) is given by:

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

Where \(h\) (Planck's constant) is approximately \(6.626 \times 10^{-34} \, \text{Js}\). Now substituting the values:

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{5.39 \times 10^{-23} \, \text{kg m/s}} \]

Calculating this gives:

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

### Step 5: Convert to Angstroms
To convert meters to angstroms (1 angstrom = \(10^{-10} \, \text{m}\)):

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

### Final Answer
The wavelength of the electron with energy 10 keV is approximately \(0.12 \, \text{Å}\).

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