The kinetic energy of an electron is `5 xx 10^(5) eV` (electron volts). Calculate the wavelength of the wave associated with the electron. The mass of the electron may be taken as `10^(-30) kg`

Explanation

To calculate the wavelength of the wave associated with an electron given its kinetic energy, we can follow these steps:

### Step 1: Convert Kinetic Energy from eV to Joules
The kinetic energy (KE) of the electron is given as \( 5 \times 10^5 \) eV. We need to convert this energy into joules using the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \).

\[ KE = 5 \times 10^5 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 8 \times 10^{-14} \text{ J} \]

### Step 2: Use the Kinetic Energy Formula to Find Velocity
The formula for kinetic energy is given by:

\[ KE = \frac{1}{2} mv^2 \]

Where:
- \( m \) is the mass of the electron, \( 10^{-30} \text{ kg} \)
- \( v \) is the velocity of the electron

Rearranging the formula to solve for \( v \):

\[ v^2 = \frac{2 \times KE}{m} \]

Substituting the values we have:

\[ v^2 = \frac{2 \times 8 \times 10^{-14} \text{ J}}{10^{-30} \text{ kg}} = \frac{16 \times 10^{-14}}{10^{-30}} = 16 \times 10^{16} \]

Taking the square root to find \( v \):

\[ v = \sqrt{16 \times 10^{16}} = 4 \times 10^8 \text{ m/s} \]

### Step 3: Calculate the Wavelength Using de Broglie's Equation
According to de Broglie's hypothesis, the wavelength \( \lambda \) associated with a particle is given by:

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

Where:
- \( h \) is Planck's constant, \( 6.63 \times 10^{-34} \text{ J s} \)
- \( m \) is the mass of the electron, \( 10^{-30} \text{ kg} \)
- \( v \) is the velocity we just calculated, \( 4 \times 10^8 \text{ m/s} \)

Substituting the values into the equation:

\[ \lambda = \frac{6.63 \times 10^{-34} \text{ J s}}{10^{-30} \text{ kg} \times 4 \times 10^8 \text{ m/s}} = \frac{6.63 \times 10^{-34}}{4 \times 10^{-22}} = 1.6575 \times 10^{-12} \text{ m} \]

### Final Answer
The wavelength of the wave associated with the electron is approximately:

\[ \lambda \approx 1.65 \times 10^{-11} \text{ m} \]

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