The de Broglie wavelength of a particle of mass 2 gram and velocity 100 `ms^(−1)` is:

Options

A. 6.63×`10^(−33)`
B. 3.31×`10^(−33)`
C. 5.31×`10^(−33)`
D. 4.31×`10^(−33)`

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 (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) 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 the Mass from grams to kilograms**:
Given mass = 2 grams.
To convert grams to kilograms, we use the conversion factor \(1 \, \text{gram} = 10^{-3} \, \text{kg}\):
\[ m = 2 \, \text{grams} \times 10^{-3} \, \text{kg/g} = 2 \times 10^{-3} \, \text{kg} \]

2. **Identify the Velocity**:
The velocity \(v\) is given as:
\[ v = 100 \, \text{m/s} \]

3. **Substitute the Values into the de Broglie Wavelength Formula**:
Now, we can substitute the values of \(h\), \(m\), and \(v\) into the formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(2 \times 10^{-3} \, \text{kg}) \times (100 \, \text{m/s})} \]

4. **Calculate the Denominator**:
First, calculate the denominator:
\[ mv = (2 \times 10^{-3} \, \text{kg}) \times (100 \, \text{m/s}) = 2 \times 10^{-1} \, \text{kg m/s} = 0.2 \, \text{kg m/s} \]

5. **Calculate the Wavelength**:
Now substitute this back into the equation for \(\lambda\):
\[ \lambda = \frac{6.626 \times 10^{-34}}{0.2} = 3.313 \times 10^{-33} \, \text{m} \]

6. **Final Result**:
Therefore, the de Broglie wavelength of the particle is:
\[ \lambda \approx 3.31 \times 10^{-33} \, \text{m} \]