The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of `100 ms^(-1)` is:

Options

A. `10^(-33)m`
B. `10^(-35)m`
C. `10^(-31)m`
D. `10^(-25)m`

Explanation

To find the de-Broglie wavelength of a particle, we can use the de-Broglie wavelength formula:

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

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the particle in kilograms,
- \(v\) is the velocity of the particle in meters per second.

### Step-by-Step Solution:

1. **Convert Mass from grams to kilograms**:
The mass of the particle is given as \(6.63 \, \text{g}\). To convert grams to kilograms, we use the conversion factor:
\[ 1 \, \text{kg} = 1000 \, \text{g} \]
Therefore,
\[ m = \frac{6.63 \, \text{g}}{1000} = 6.63 \times 10^{-3} \, \text{kg} \]

2. **Identify the velocity**:
The velocity \(v\) is given as \(100 \, \text{ms}^{-1}\).

3. **Substitute values into the de-Broglie wavelength formula**:
Now we can substitute the values of \(h\), \(m\), and \(v\) into the formula:
\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{(6.63 \times 10^{-3} \, \text{kg})(100 \, \text{ms}^{-1})} \]

4. **Calculate the denominator**:
First, calculate the denominator:
\[ mv = (6.63 \times 10^{-3} \, \text{kg})(100 \, \text{ms}^{-1}) = 6.63 \times 10^{-1} \, \text{kg m/s} \]

5. **Calculate the wavelength**:
Now substitute the denominator back into the equation:
\[ \lambda = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-1}} = 10^{-33} \, \text{m} \]

### Final Answer:
The de-Broglie wavelength of the particle is:
\[ \lambda = 10^{-33} \, \text{m} \]