What is the de Broglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ` " ms"^(-1)` ? What can we infer from this result ?

Options

A. `3.5xx10^(-34) m`
B. `2.8xx10^(-34) m`
C. `1.2xx10^(-34) m`
D. `2.1xx10^(-34) m`

Answer

To find the de Broglie wavelength of a ball with a mass of 0.12 kg moving at a speed of 20 m/s, we can use the de Broglie wavelength formula:

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

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is the Planck constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the particle (0.12 kg),
- \(v\) is the velocity of the particle (20 m/s).

### Step 1: Identify the values
- Mass \(m = 0.12 \, \text{kg}\)
- Velocity \(v = 20 \, \text{ms}^{-1}\)
- Planck constant \(h = 6.63 \times 10^{-34} \, \text{Js}\)

### Step 2: Calculate the momentum
First, we calculate the momentum \(mv\):
\[ mv = 0.12 \, \text{kg} \times 20 \, \text{ms}^{-1} = 2.4 \, \text{kg m/s} \]

### Step 3: Substitute values into the de Broglie wavelength formula
Now, we substitute the values into the de Broglie wavelength formula:
\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{2.4 \, \text{kg m/s}} \]

### Step 4: Perform the division
Calculating the above expression:
\[ \lambda = \frac{6.63 \times 10^{-34}}{2.4} \approx 2.7625 \times 10^{-34} \, \text{m} \]

### Step 5: Round off the result
Rounding off to two significant figures, we get:
\[ \lambda \approx 2.8 \times 10^{-34} \, \text{m} \]

### Conclusion
The de Broglie wavelength of the ball is approximately \(2.8 \times 10^{-34} \, \text{m}\).

### Inference
From this result, we can infer that the de Broglie wavelength of macroscopic objects (like a ball) is extremely small, indicating that wave-like properties are negligible for such objects. This suggests that classical mechanics is sufficient to describe their motion, as quantum effects are not observable at this scale.

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