What will be the wavelength of an electron moving with 1/10th of velocity of light?

Options

A. `2.43xx10^(-11)`m
B. `243xx10^(-11)`m
C. 0.243 m
D. `2.43xx10^(-4)`m

Explanation

To find the wavelength of an electron moving at 1/10th of the velocity of light, we can use the de Broglie wavelength formula:

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

Where:
- \(\lambda\) is the wavelength,
- \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(p\) is the momentum of the electron.

### Step 1: Calculate the momentum of the electron

The momentum \(p\) of an object is given by the formula:

\[ p = mv \]

Where:
- \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)),
- \(v\) is the velocity of the electron.

Given that the electron is moving at \(1/10\)th of the velocity of light, we can express this as:

\[ v = \frac{c}{10} \]

Where \(c\) (the speed of light) is approximately \(3.00 \times 10^8 \, \text{m/s}\).

### Step 2: Substitute the values into the momentum formula

Substituting the values into the momentum formula:

\[ v = \frac{3.00 \times 10^8 \, \text{m/s}}{10} = 3.00 \times 10^7 \, \text{m/s} \]

Now, calculate the momentum:

\[ p = m \cdot v = (9.11 \times 10^{-31} \, \text{kg}) \cdot (3.00 \times 10^7 \, \text{m/s}) \]

Calculating this gives:

\[ p = 2.733 \times 10^{-23} \, \text{kg m/s} \]

### Step 3: Calculate the wavelength using the de Broglie formula

Now, we can substitute the momentum back into the de Broglie wavelength formula:

\[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \, \text{Js}}{2.733 \times 10^{-23} \, \text{kg m/s}} \]

Calculating this gives:

\[ \lambda \approx 2.43 \times 10^{-11} \, \text{m} \]

### Final Answer

Thus, the wavelength of an electron moving with \(1/10\)th of the velocity of light is approximately:

\[ \lambda \approx 2.43 \times 10^{-11} \, \text{m} \]