The de-Broglie wavelength of an electron is `600 nm`. The velocity of the electron is:
`(h = 6.6 xx 10^(-34) J "sec", m = 9.0 xx 10^(-31) kg)`

Options

A. `1.8 xx 10^3 m s^(-1)`
B. `1.2 xx 10^(5) ms^(-1)`
C. `5.4 xx 10^3 ms^(-1)`
D. `1.2 xx 10^3 ms^(-1)`

Explanation

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

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

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is the Planck constant,
- \(p\) is the momentum of the electron.

The momentum \(p\) can be expressed as:

\[ p = mv \]

where:
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron.

Substituting the expression for momentum into the de-Broglie wavelength formula gives us:

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

Rearranging this equation to solve for velocity \(v\):

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

Now, we can plug in the values provided in the question:

1. **Planck constant \(h\)**: \(6.6 \times 10^{-34} \, \text{J sec}\)
2. **Mass of electron \(m\)**: \(9.0 \times 10^{-31} \, \text{kg}\)
3. **Wavelength \(\lambda\)**: \(600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}\)

Now substituting these values into the equation:

\[ v = \frac{6.6 \times 10^{-34} \, \text{J sec}}{(9.0 \times 10^{-31} \, \text{kg})(600 \times 10^{-9} \, \text{m})} \]

Calculating the denominator:

\[ 9.0 \times 10^{-31} \, \text{kg} \times 600 \times 10^{-9} \, \text{m} = 5.4 \times 10^{-28} \, \text{kg m} \]

Now substituting this back into the equation for velocity:

\[ v = \frac{6.6 \times 10^{-34}}{5.4 \times 10^{-28}} \]

Calculating this gives:

\[ v \approx 1.222 \times 10^{4} \, \text{m/s} \approx 1.22 \times 10^{3} \, \text{m/s} \]

Thus, the velocity of the electron is approximately:

\[ \boxed{1.22 \times 10^{3} \, \text{m/s}} \]