The de Broglie wavelenth of `1 mg` grain of sand blown by a `20 m s^-1` wind is :

Options

A. `3.3 xx 10^-29 m`
B. `3.3 xx 10^-21 m`
C. `3.3 xx 10^-49 m`
D. `3.3 xx 10^-42 m`

Explanation

To find the de Broglie wavelength of a 1 mg grain of sand blown by a 20 m/s wind, we can use the de Broglie wavelength formula:

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

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the particle (in kg),
- \(v\) is the velocity of the particle (in m/s).

### Step 1: Convert the mass from mg to kg
Given that the mass of the grain of sand is 1 mg, we need to convert this to kilograms:

\[ 1 \, \text{mg} = 1 \times 10^{-3} \, \text{g} = 1 \times 10^{-6} \, \text{kg} \]

### Step 2: Identify the velocity
The velocity of the grain of sand is given as:

\[ v = 20 \, \text{m/s} \]

### Step 3: Substitute values into the de Broglie wavelength formula
Now we can substitute the values of \(h\), \(m\), and \(v\) into the de Broglie wavelength formula:

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(1 \times 10^{-6} \, \text{kg}) \times (20 \, \text{m/s})} \]

### Step 4: Calculate the denominator
First, calculate the denominator:

\[ m \cdot v = (1 \times 10^{-6} \, \text{kg}) \cdot (20 \, \text{m/s}) = 2 \times 10^{-5} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie wavelength
Now substitute this back into the equation for \(\lambda\):

\[ \lambda = \frac{6.626 \times 10^{-34}}{2 \times 10^{-5}} = 3.313 \times 10^{-29} \, \text{m} \]

### Final Answer
The de Broglie wavelength of a 1 mg grain of sand blown by a 20 m/s wind is:

\[ \lambda = 3.313 \times 10^{-29} \, \text{m} \]

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