A 2 mg sand particle is blown with a speed of `50 m//sec` what is the its de Broglie's wavelength ?

Options

A. `3.31xx10^(-30)m`
B. `3.31xx10^(-15)m`
C. `6.62xx10^(-30)m`
D. `6.62xx10^(-15)m`

Explanation

To find the de Broglie wavelength of a 2 mg sand particle moving at a speed of 50 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) = Planck's constant = \( 6.63 \times 10^{-34} \, \text{Joule second} \)
- \( m \) = mass of the particle in kg
- \( v \) = velocity of the particle in m/s

2. **Convert Mass to Kilograms**:
The mass of the sand particle is given as 2 mg. We need to convert this to kilograms:
\[ m = 2 \, \text{mg} = 2 \times 10^{-3} \, \text{g} = 2 \times 10^{-6} \, \text{kg} \]

3. **Identify the Velocity**:
The velocity \( v \) is given directly as:
\[ v = 50 \, \text{m/s} \]

4. **Substitute Values into the Formula**:
Now, we can substitute the values of \( h \), \( m \), and \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{6.63 \times 10^{-34}}{(2 \times 10^{-6}) \times (50)} \]

5. **Calculate the Denominator**:
First, calculate the denominator:
\[ (2 \times 10^{-6}) \times (50) = 1 \times 10^{-4} \, \text{kg m/s} \]

6. **Perform the Division**:
Now, perform the division:
\[ \lambda = \frac{6.63 \times 10^{-34}}{1 \times 10^{-4}} = 6.63 \times 10^{-30} \, \text{m} \]

7. **Final Result**:
The de Broglie wavelength of the sand particle is:
\[ \lambda = 6.63 \times 10^{-30} \, \text{m} \]