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Chapter 36: Problem 40

What is the de Broglie wavelength of a \(2.000 \cdot 10^{3}-\mathrm{kg}\) car moving at a speed of \(100.0 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
Answer: The de Broglie wavelength of the car is approximately 1.19 × 10⁻³⁸ meters.

Step by step solution

01

Convert speed from km/h to m/s

To convert the speed of the car from 100.0 km/h to m/s, we need to multiply it by a conversion factor. There are 1000 meters in a kilometer and 3600 seconds in an hour, so the conversion factor is: \(\text{conversion factor} = \dfrac{1000 \text{ meters}}{1 \text{ kilometer}} \times \dfrac{1 \text{ hour}}{3600 \text{ seconds}}\) Now, we can convert the speed: \(v = 100.0 \dfrac{\text{km}}{\text{h}} \times \text{conversion factor} = 100.0 \dfrac{\text{km}}{\text{h}} \times \dfrac{1000 \text{ meters}}{1 \text{ kilometer}} \times \dfrac{1 \text{ hour}}{3600 \text{ seconds}} = 27.8\ \dfrac{\text{m}}{\text{s}}\)
02

Calculate momentum

Now that we have the speed of the car in m/s, we can calculate its momentum using the formula: \(p = mv\) where \(m = 2.000 \times 10^{3}\ \text{kg}\) (mass of the car) and \(v = 27.8\ \dfrac{\text{m}}{\text{s}}\) (speed of the car): \(p = (2.000 \times 10^{3}\ \text{kg})(27.8\ \dfrac{\text{m}}{\text{s}}) = 5.56 \times 10^{4}\ \text{kg}\cdot \text{m} / \text{s}\)
03

Compute the de Broglie wavelength

Finally, we can use the formula, \(\lambda = \dfrac{h}{p}\), to calculate the de Broglie wavelength. Planck's constant, \(h\), is approximately \(6.63 \times 10^{-34}\ \text{J}\cdot\text{s}\). We can rewrite Planck's constant in units of \(\text{kg}\cdot\text{m}^2/\text{s}\) (as \(1\ \text{J} = 1\ \text{kg}\cdot\text{m}^2/\text{s}^2\)): \(h = 6.63 \times 10^{-34}\ \text{kg}\cdot\text{m}^2/\text{s}\) Now, we can find the de Broglie wavelength: \(\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}\ \text{kg}\cdot\text{m}^2/\text{s}}{5.56 \times 10^{4}\ \text{kg}\cdot\text{m}/\text{s}} = 1.19 \times 10^{-38}\ \text{m}\) The de Broglie wavelength of the car is approximately \(1.19 \times 10^{-38}\) meters.

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Most popular questions from this chapter

A nocturnal bird's eye can detect monochromatic light of frequency \(5.8 \cdot 10^{14} \mathrm{~Hz}\) with a power as small as \(2.333 \cdot 10^{-17} \mathrm{~W}\). What is the corresponding number of photons per second a nocturnal bird's eye can detect?

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