Question

The de-Broglie wavelength associated with a ball of mass 150 g traveling at 30.0 m/s would be

• A. 1.47 x 10\(^{-34}\) m
• B. 14.7 x 10\(^{-34}\) m
• C. 0.147 x 10\(^{-34}\) m
• D. 7.14 x 10\(^{-34}\) m

Hint:

For macroscopic objects (like a ball), the de-Broglie wavelength is extremely small, typically on the order of \(10^{-34}\) m. This is far too small to be measured, which is why we don't observe wave-like properties for everyday objects. Always convert mass to kilograms before calculation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires the calculation of the de-Broglie wavelength of a macroscopic object. The de-Broglie hypothesis states that all matter has wave-like properties, and the wavelength (\(\lambda\)) is inversely proportional to the momentum (\(p\)) of the object.

Step 2: Key Formula or Approach:
The de-Broglie wavelength (\(\lambda\)) is given by the formula: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\)), \(m\) is the mass of the object, and \(v\) is its velocity.

Step 3: Detailed Explanation:
Given data:
Mass, \(m = 150 \, \text{g} = 0.150 \, \text{kg}\) (It's crucial to convert to SI units).
Velocity, \(v = 30.0 \, \text{m/s}\).
Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\).
Calculation:
First, calculate the momentum \(p\): \[ p = mv = (0.150 \, \text{kg}) \times (30.0 \, \text{m/s}) = 4.5 \, \text{kg}\cdot\text{m/s} \] Now, calculate the de-Broglie wavelength: \[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \, \text{J}\cdot\text{s}}{4.5 \, \text{kg}\cdot\text{m/s}} \] \[ \lambda \approx 1.4724 \times 10^{-34} \, \text{m} \] Rounding the result, we get \(1.47 \times 10^{-34} \, \text{m}\).

Step 4: Final Answer:
The de-Broglie wavelength of the ball is \(1.47 \times 10^{-34}\) m.