The de-Broglie wavelength of an electron is 66 nm. The velocity of the electron is `[h= 6.6 xx 10^(-34) kg m^(2)s^(-1), m=9.0 xx 10^(-31) kg]`

Options

A. `1.84 xx 10^(-4) ms^(-1)`
B. `1.1 xx 10^(4) ms^(-1)`
C. `5.4 xx 10^(3) ms^(-1)`
D. `1.1 xx 10^(3) ms^(-1)`

Answer

To find the velocity of the electron given its de-Broglie wavelength, we can use the de-Broglie wavelength formula:

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

Where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is Planck's constant,
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron.

### Step 1: Convert the Wavelength to Meters
The given wavelength is \(66 \, \text{nm}\). We need to convert this to meters:

\[ \lambda = 66 \, \text{nm} = 66 \times 10^{-9} \, \text{m} \]

### Step 2: Rearrange the Formula to Solve for Velocity
We need to find the velocity \(v\). Rearranging the formula gives us:

\[ v = \frac{h}{m\lambda} \]

### Step 3: Substitute the Known Values
Now, substitute the known values into the formula:

- \(h = 6.6 \times 10^{-34} \, \text{kg m}^2/\text{s}\)
- \(m = 9.0 \times 10^{-31} \, \text{kg}\)
- \(\lambda = 66 \times 10^{-9} \, \text{m}\)

Substituting these values in:

\[ v = \frac{6.6 \times 10^{-34}}{(9.0 \times 10^{-31})(66 \times 10^{-9})} \]

### Step 4: Calculate the Denominator
Calculate the denominator:

\[ 9.0 \times 10^{-31} \times 66 \times 10^{-9} = 594 \times 10^{-40} = 5.94 \times 10^{-38} \]

### Step 5: Calculate the Velocity
Now substitute the denominator back into the equation for velocity:

\[ v = \frac{6.6 \times 10^{-34}}{5.94 \times 10^{-38}} \]

Calculating this gives:

\[ v \approx 1.11 \times 10^{4} \, \text{m/s} \]

### Final Answer
The velocity of the electron is approximately:

\[ v \approx 1.11 \times 10^{4} \, \text{m/s} \]