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Chapter 6: Problem 47

Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an \(85-\mathrm{kg}\) person skiing at $50 \mathrm{~km} / \mathrm{hr},(\mathbf{b})\( a \)10.0-\mathrm{g}\( bullet fired at \)250 \mathrm{~m} / \mathrm{s},(\mathbf{c})\( a lithium atom moving at \)2.5 \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathbf{d})\( an ozone \)\left(\mathrm{O}_{3}\right)$ molecule in the upper atmosphere moving at \(550 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The wavelengths of the objects are as follows: a) \(5.61 \times 10^{-37} m\) b) \(2.65 \times 10^{-34} m\) c) \(3.82 \times 10^{-13} m\) d) \(1.52 \times 10^{-10} m\)

Step by step solution

01

Convert velocities into a suitable unit

First, we need to convert the velocities of moving objects into meters per second (m/s) if they are given in a different unit, such as km/hr. a) Convert 50 km/hr to m/s \[ 50 \frac{km}{hr} \times \frac{1000 m}{1 km} \times \frac{1 hr}{3600 s} = \frac{50 \times 1000}{3600} m/s \approx 13.89 m/s \] Now we have the velocities in m/s.
02

Calculate the momentum for each object

We can use the below formula to calculate the momentum. \[ p = mv \] a) The person skiing p1 = (85 kg)(13.89 m/s) ≈ 1180.65 kg m/s b) The bullet p2 = (0.01 kg)(250 m/s) = 2.5 kg m/s (note: 10 g was converted to kg) c) The lithium atom (mass = 6.939 * 10^(-27) kg) p3 = (6.939 * 10^(-27) kg)(2.5 * 10^5 m/s) ≈ 1.735 * 10^(-21) kg m/s d) The ozone molecule (mass = 3 * 16 * 1.661 * 10^(-27) kg, since it has 3 oxygen atoms) p4 = (3 * 16 * 1.661 * 10^(-27) kg)(550 m/s) ≈ 4.358 * 10^(-24) kg m/s
03

Find the wavelengths using de Broglie relationship

Now we can find the wavelengths using the de Broglie relationship formula. \[\lambda = \frac{h}{p}\] a) The person skiing λ1 = (6.626 * 10^(-34) Js) / (1180.65 kg m/s) ≈ 5.61 * 10^(-37) m b) The bullet λ2 = (6.626 * 10^(-34) Js) / (2.5 kg m/s) ≈ 2.65 * 10^(-34) m c) The lithium atom λ3 = (6.626 * 10^(-34) Js) / (1.735 * 10^(-21) kg m/s) ≈ 3.82 * 10^(-13) m d) The ozone molecule λ4 = (6.626 * 10^(-34) Js) / (4.358 * 10^(-24) kg m/s) ≈ 1.52 * 10^(-10) m So, the wavelengths of the objects are as follows: a) 5.61 * 10^(-37) m b) 2.65 * 10^(-34) m c) 3.82 * 10^(-13) m d) 1.52 * 10^(-10) m

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