The de Broglie wavelength of a particle of mass 1 gram and velocity 300 `ms^(−1)` is:

Options

A. 3.63×`10^(−33)`
B. 2.20×`10^(−33)`
C. 4.63×`10^(−33)`
D. 6.63×`10^(−33)`

Answer

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:

1. **Convert the mass from grams to kilograms:**
\[ m = 1 \, \text{gram} = 1 \times 10^{-3} \, \text{kg} \]

2. **Identify the velocity of the particle:**
\[ v = 300 \, \text{m/s} \]

3. **Substitute the values into the de Broglie wavelength formula:**
\[ \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \, \text{Js}}{(1 \times 10^{-3} \, \text{kg}) \times (300 \, \text{m/s})} \]

4. **Calculate the denominator:**
\[ mv = (1 \times 10^{-3} \, \text{kg}) \times (300 \, \text{m/s}) = 3 \times 10^{-1} \, \text{kg m/s} \]

5. **Now, substitute the denominator back into the equation:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{3 \times 10^{-1}} \]

6. **Perform the division:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{3} = 2.20867 \times 10^{-34} \, \text{m} \]

7. **Round the result to two decimal places:**
\[ \lambda \approx 2.21 \times 10^{-34} \, \text{m} \]

### Final Result:
The de Broglie wavelength of the particle is approximately \(2.21 \times 10^{-34} \, \text{m}\).