If the de Broglie wavelength for a 155 g hockey puck is 1.14 × 10–34 m, what is the velocity (in m/s) of the puck?

Round your answer to ONE decimal place (i.e. 2.7) and do NOT include the units of m/s.
If your last digit is a zero after a decimal (i.e. 6.40), Carmen may drop the zero. That is OK!

15A

Question

If the de Broglie wavelength for a 155 g hockey puck is 1.14 × 10–34 m, what is the velocity (in m/s) of the puck?

Round your answer to ONE decimal place (i.e. 2.7) and do NOT include the units of m/s.
If your last digit is a zero after a decimal (i.e. 6.40), Carmen may drop the zero. That is OK!

15A

BlackTom AI · Accepted Answer

We start from the de Broglie relation λ = h / p, where p is the momentum. Since p = m v, we can solve for velocity: v = h / (m λ).

First, convert the given mass to kilograms: 155 g = 0.155 kg.
Next, note the provided de Broglie wavelength: λ = 1.14 × 10^−34 m.
Use Planck’s constant: h ≈ 6.62607015 × 10^−34 J·s.
Compute the denominator: m × λ = 0.155 kg × 1.14 × 10^−34 m = 0.1767 × 10^−34 kg·m = 1.767 × 10^−35 kg·m.
Now divide h by this product: v = h / (m λ) = (6.62607015 × 10^−34) / (1.767 × 10^−35) ≈ 37.5.
Since the problem asks for the value without units and to one decimal place, the velocity is 37.5. The units in the calculation cancel appropriately to yield meters per second as expected.

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