Calculate the de-Broglie wavelength of an electron of kinetic energy 100 eV. Given `m_(e)=9.1xx10^(-31)kg, h=6.62xx10^(-34)Js`.

Explanation

To calculate the de-Broglie wavelength of an electron with a kinetic energy of 100 eV, we can follow these steps:

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

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

### Step 2: Calculate the Momentum of the Electron
The kinetic energy can also be expressed in terms of momentum (p) as:
\[ KE = \frac{p^2}{2m} \]
Rearranging this gives:
\[ p = \sqrt{2m \cdot KE} \]

Substituting the mass of the electron \( m = 9.1 \times 10^{-31} \text{ kg} \) and the kinetic energy we just calculated:
\[ p = \sqrt{2 \times (9.1 \times 10^{-31} \text{ kg}) \times (1.6 \times 10^{-17} \text{ J})} \]

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

### Step 3: Use the de-Broglie Wavelength Formula
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h = 6.62 \times 10^{-34} \text{ Js} \).

Substituting the value of \( p \):
\[ \lambda = \frac{6.62 \times 10^{-34} \text{ Js}}{5.39 \times 10^{-24} \text{ kg m/s}} \]

Calculating this gives:
\[ \lambda \approx 1.227 \times 10^{-10} \text{ m} \]

### Final Answer
The de-Broglie wavelength of the electron is approximately:
\[ \lambda \approx 1.227 \times 10^{-10} \text{ m} \]

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