Question:medium

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

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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.
Updated On: Mar 27, 2026
  • 1.47 x 10\(^{-34}\) m
  • 14.7 x 10\(^{-34}\) m
  • 0.147 x 10\(^{-34}\) m
  • 7.14 x 10\(^{-34}\) m
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The Correct Option is A

Solution and Explanation


Step 1: Conceptual Foundation:
This problem necessitates the computation of the de-Broglie wavelength for a macroscopic object. The de-Broglie hypothesis posits that all matter exhibits wave-like characteristics, with wavelength (\(\lambda\)) being inversely proportional to the object's momentum (\(p\)).

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

Step 3: Procedural Breakdown:
Provided Data:
Mass, \(m = 150 \, \text{g} = 0.150 \, \text{kg}\) (Conversion to SI units is mandatory).
Velocity, \(v = 30.0 \, \text{m/s}\).
Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\). Calculations:
Initially, calculate the momentum \(p\): \[ p = mv = (0.150 \, \text{kg}) \times (30.0 \, \text{m/s}) = 4.5 \, \text{kg}\cdot\text{m/s} \] Subsequently, compute the de-Broglie wavelength: \[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \, \text{J}\cdot\text{s}}{4.5 \, \text{kg}\cdot\text{m/s}} \] \[ \lambda \approx 1.4724 \times 10^{-34} \, \text{m} \] Upon rounding, the result is \(1.47 \times 10^{-34} \, \text{m}\).

Step 4: Conclusive Result:
The de-Broglie wavelength of the ball is determined to be \(1.47 \times 10^{-34}\) m.

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