Problem 47 - Chemistry: The Central Science in SI Units, Global Edition

Question

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 \(550 \mathrm{~m} / \mathrm{s}\).

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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

Step 3: 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

Key Concepts

Momentum Calculation

Momentum is a key concept in physics. When dealing with moving objects, we often need to calculate their momentum to understand their behavior better. Momentum essentially measures how hard it is to stop a moving object. It's a product of an object's mass and velocity. The formula is simple: \[ p = mv \]where:

• \( p \) is the momentum,
• \( m \) is the mass of the object,
• \( v \) is the velocity of the object.

For example, when calculating the momentum of a person skiing or a bullet being fired, we multiply their mass by their respective speeds. Remember, momentum has units of kilogram meters per second \( \text{(kg m/s)} \). Understanding momentum is crucial because it provides a bridge to several fundamental concepts in quantum mechanics, such as de Broglie wavelengths used to describe the wave nature of matter.

Wave-Particle Duality

Wave-particle duality is a fascinating principle of quantum mechanics. It suggests that every particle or quantum entity can exhibit both wave-like and particle-like properties. This dual nature is not something we encounter in everyday life, making it challenging to grasp initially.For particles like electrons or even larger objects, this wave aspect is described using the de Broglie wavelength. The de Broglie relationship helps us calculate the wavelength \( (\lambda) \) for a moving particle via:\[ \lambda = \frac{h}{p} \]This means that even large objects, like a person skiing, have a wave nature, albeit with an incredibly small wavelength, making it practically impossible to observe. However, for very small particles such as electrons, this wavelength becomes significant, drastically affecting how they behave and interact with their environment.

Unit Conversion

Unit conversion is a fundamental skill in physics that ensures coherence and accuracy. When solving problems, you often need units to be compatible with each other. For example, velocity is often given in kilometers per hour \( (\text{km/hr}) \), but when calculating momentum or wave properties, it’s needed in meters per second \( (\text{m/s}) \).To convert from \( \text{km/hr} \) to \( \text{m/s} \), you can use the following conversion factor:\[ 1 \text{ km/hr} = \frac{1000 \text{ meters}}{3600 \text{ seconds}} \approx 0.278 \text{ m/s} \]Thus, converting helps maintain consistency across calculations, ensuring values like momentum can be calculated accurately. This skill is not just about plugging numbers into a formula but understanding the relationship between different units, which is crucial in making sense of your results.

Planck's Constant

Planck's constant \( (h) \) is a vital component in quantum mechanics. It is a fundamental constant used to describe the sizes of quanta. The value is approximately:\[ h \approx 6.626 \times 10^{-34} \text{ Js} \]It appears in many quantum equations, with the de Broglie wavelength formula being one of the most notable. This constant helps bridge the gap between the wave and particle descriptions of light and matter.Planck's constant essentially allows scientists to calculate the energy carried by photons or the energy levels of atoms. It's a pivotal element in equations that describe how tiny particles behave, providing insights into the quantum world that defy classical physics' laws. Understanding Planck's constant gives you a clearer view of the structure of the universe at its most fundamental level.