The de Broglie wavelength of a particle of mass 1 gram and velocity 200 `ms^(−1)` is:

Options

A. 6.63×`10^(−33)`
B. 3.31×`10^(−33)`
C. 4.63×`10^(−33)`
D. 5.63×`10^(−33)`

Explanation

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

**Step 1: Convert the mass from grams to kilograms.**
- Given mass \(m = 1 \, \text{g}\).
- To convert grams to kilograms, we divide by 1000:
\[ m = 1 \, \text{g} = \frac{1}{1000} \, \text{kg} = 0.001 \, \text{kg} \]

**Step 2: Identify the velocity.**
- Given velocity \(v = 200 \, \text{ms}^{-1}\).

**Step 3: Substitute the values into the de Broglie wavelength formula.**
- Using the values \(h = 6.626 \times 10^{-34} \, \text{Js}\), \(m = 0.001 \, \text{kg}\), and \(v = 200 \, \text{ms}^{-1}\):
\[ \lambda = \frac{6.626 \times 10^{-34}}{0.001 \times 200} \]

**Step 4: Calculate the denominator.**
- Calculate \(mv\):
\[ mv = 0.001 \, \text{kg} \times 200 \, \text{ms}^{-1} = 0.2 \, \text{kg m/s} \]

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

**Step 6: Round the answer.**
- The final answer can be rounded to:
\[ \lambda \approx 3.31 \times 10^{-33} \, \text{m} \]

### Final Answer:
The de Broglie wavelength of the particle is approximately \(3.31 \times 10^{-33} \, \text{m}\).

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